HA150
SQL Basics for SAP HANA
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INSTRUCTOR HANDBOOK
INSTRUCTOR-LED TRAINING
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Course Version: 12
Course Duration: 2 Day(s)
Material Number: 50135954
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American English is the standard used in this handbook.
The following typographic conventions are also used.
This information is displayed in the instructor’s presentation
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Contents
vii
Course Overview
1
Unit 1:
2
15
21
Lesson: Understanding Motivation and Basic Concepts
Lesson: Understanding the Sample Data
Unit 2:
22
47
53
Unit 3:
70
91
95
96
107
111
112
117
121
122
129
133
134
143
149
150
157
30 Minutes
15 Minutes
Data Storage
Lesson: Defining How Data is Stored
Exercise 7: Create, Change, and Delete a Database Table
Unit 9:
30 Minutes
15 Minutes
Data Stored in Tables
Lesson: Changing Data Stored in Tables
Exercise 6: Insert, Update, and Delete Rows in a Database Table
Unit 8:
40 Minutes
10 Minutes
NULL Values
Lesson: Understanding NULL Values
Exercise 5: Understand NULL Values
Unit 7:
60 Minutes
20 Minutes
Data From Multiple Tables Part II
Lesson: Reading Data from Multiple Tables- Part 2
Exercise 4: Use Nested Queries
Unit 6:
45 Minutes
20 Minutes
Data From Multiple Tables Part I
Lesson: Reading Data from Multiple Tables- Part 1
Exercise 3: Use JOIN to Combine Data from Several Tables
Unit 5:
75 Minutes
25 Minutes
Aggregation of Data
Lesson: Aggregating Data
Exercise 2: Use Aggregate Expressions
Unit 4:
25 Minutes
10 Minutes
Data From a Table Or View
Lesson: Using Data from a Table or View
Exercise 1: Using the SELECT Statement to Read from a Single
Table
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69
Motivation and Basic Concepts
45 Minutes
15 Minutes
Views For Data Access
Lesson: Using Views for Data Access
Exercise 8: Work with Database Views
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30 Minutes
10 Minutes
v
161
Unit 10:
162
167
173
Lesson: Creating User-Defined Functions
Exercise 9: Create and Call Scalar User-Defined Functions
Exercise 10: Create and Call Table User-Defined Functions
80 Minutes
15 Minutes
30 Minutes
178
185
Lesson: Creating Database Procedures
Exercise 11: Create and Use a Database Procedure
60 Minutes
25 Minutes
191
Unit 11:
192
201
202
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Functions and Procedures
Data Access
Lesson: Defining Data Access
Unit 12:
20 Minutes
Database Transactions
Lesson: Explaining Database Transactions
25 Minutes
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Course Overview
TARGET AUDIENCE
This course is intended for the following audiences:
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Application Consultant
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Development Consultant
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Technology Consultant
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UNIT 1
Motivation and Basic Concepts
Lesson 1
Understanding Motivation and Basic Concepts
2
Lesson 2
Understanding the Sample Data
15
UNIT OBJECTIVES
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Understand the motivation for and foundation of the relational model
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Understand SQL and its relation to the relation model
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Understand database tables as the most important database objects
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Understand the sample database used throughout the course
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Unit 1
Lesson 1
2
Understanding Motivation and Basic Concepts
LESSON OVERVIEW
This lessons explains why learning SAP HANA SQL may be important and the relational
database model it originates from.
LESSON OBJECTIVES
After completing this lesson, you will be able to:
●
Understand the motivation for and foundation of the relational model
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Understand SQL and its relation to the relation model
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Understand database tables as the most important database objects
Terms Database Management System, Database and Database System
The terms Database, Database System and Database Management System are frequently
used incorrectly and interchangeably:
Figure 1: Terms DB, DBS, and DBMS
The terms in fact refer to slightly different concepts:
Database (DB)
A database is a concrete structured collection of “records”.
Database Management System (DBMS)
A database management system is software that manages databases. Every access to a
database (create, read, insert, update, delete) goes exclusively through the DBMS. The
DBMS exercises complete control over the database(s) it manages.
Database System
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Lesson: Understanding Motivation and Basic Concepts
A database system is the combination of the implementation of a DBMS and the
database(s) it manages.
Figure 2: DBS = DB + DBMS
Goals of the Relational Database Model
Several database models, that are ways structure data records and thus implement DBMSs,
have been developed during the last decades. The following list is a selection of such models:
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Hierarchical Database Model
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Network Database Model
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Relational Database Model
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Object-relational Database Model
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Object-oriented Database Model
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XML-based Database Model
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....
Design Goals of the Relational Database Model
The Relational Database Model was invented in the late 1960s by Edgar F. Codd (1923–2003),
then working at IBM Aldaden Research Lab in San José, California, with the following specific
goals:
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Provide a simple yet mathematically profound and precise database model
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Make monitoring data integrity largely a responsibility of the DBMS
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Separate the conceptual schema from the internal schema
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Make data storage and retrieval the sole responsibility of the DBMS
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A descriptive data access language removes need to traverse data links
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Database optimizers choose the optimal execution strategy for a given query
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Unit 1: Motivation and Basic Concepts
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Provide a simple database access language with semantics described with mathematical
precision
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Easy to learn
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Equivalence of different queries can be proved
The goal behind separating the conceptual schema from the internal schema is to allow for a
3–level architecture such that changes on a lower level do not affect higher levels.
Figure 3: Three Level Schema Architecture
External Schema: How is data presented to the user?
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(Partial) views of the data as required by applications or users.
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In case of a relational database system implemented via views.
Conceptual Schema: What is stored?
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Overall presentation of the data model at the logical, (if possible) DBMS and application
independent level.
For example, in relational representation, or even higher level of abstraction (e.g. E/R
model).
Internal Schema: How and where stored?
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4
Describes (DBMS specific) the internal, physical representation of data: How and where
exactly the data is stored, internal record format, access paths, etc.
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Lesson: Understanding Motivation and Basic Concepts
Figure 4: Three Level Schema Architecture Overview
Why the Term “Relational” Database Model?
The database model is called relational because it is based on the notion of relation in set
theory. A relation in the mathematical sense is a subset of the Cartesian product of other sets.
Figure 5: Relations
Figure 6: Simple Base Set and Cartesian Example
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Unit 1: Motivation and Basic Concepts
Figure 7: Comparative Relation <= Example
Figure 8: Base Sets with Business Data Example
Figure 9: Employee Records as a Relation Example
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Lesson: Understanding Motivation and Basic Concepts
Figure 10: Relations and Tables
The Role of SQL
Languages for the Relational Database Model
The following languages are available to interact with a relational DBMS:
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Relational Algebra
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6 basic operations: Selection, Projection, Cartesian Product, Union, Difference and
Rename
Relational Calculus
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Formal basis for DBMS internal query optimization
Tuple variables and quantifiers
SQL
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Standardized and used in practice
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Unit 1: Motivation and Basic Concepts
Figure 11: Why SQL?
ANSI SQL
SQL is a computer language that is used to define, manipulate and control relational
databases. It is defined by the American Standards Institute (ANSI).
SQL is designed as a descriptive rather than procedural language. Using SQL you express
what data you want, not exactly how to retrieve it. This design allows the DBMS to first parse
each statement and then optimize it. The optimizer determines the best execution plan (at
least in theory).
SQL Language Elements
Table 1: SQL Language Elements
SQL language elements can be divided into the following three categories:
Data Manipulation Language
(DML) statements
DML statements insert, update and delete the data in
a database. DML statements also select data that you
want to read.
Data Definition Language
(DDL) statements
DDL statements create, alter and drop databases, tables and other database objects in the server.
Data Control Language (DCL)
statements
DCL statements grant and revoke permissions from database users.
SQL is not Fully Relational
SQL is the most important database language for the relational database model. But it
deviates in important points from the “purely relational” model:
Since SQL is designed to work with data stored in tables, in other words with sets, it is multiset oriented and not single record-based. Using a single SQL statement, you can read multiple
table rows, modify or delete them in one go.
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Lesson: Understanding Motivation and Basic Concepts
SQL is not Fully Relational
●
Results of SQL queries can contain duplicates (identical rows). This makes SQL multi-set
oriented and not set oriented.
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SQL allows for NULL values, leading to a three-valued logic.
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The SQL language is not closed.
SAP HANA SQL and SQLScript
SAP HANA SQL, including SQLScript, is the computer language created by SAP and used in
the SAP HANA platform. It is built on a foundation of standard ANSI SQL, and it extends the
standard considerably, in part to better support the column and in-memory store nature of
SAP HANA.
SAP HANA SQL and SQLScript Elements
The following is a list of selected elements of SAP HANA SQL:
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Standard (ANSI) SQL statements
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SAP’s extensions to SQL statements for SAP HANA, e.g.
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DML extentsions, e.g. SELECT INTO, UPSERT,
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DDL extensions, e.g. to allow creating column and row store tables
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Functions
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Triggers
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Declarative SQLScript statements, for example variable assignments
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Imperative SQLScript statements (flow control statements)
Database Objects
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Tables are the primary database object, but not the only one. Apart from tables a database
usually contains the following:
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Views to simplify and limit data access
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Indexes to speed up (certain) read accesses
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Constraints to ensure data consistency
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Stored procedures for more complex tasks
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Triggers to selectively respond to particular events
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Unit 1: Motivation and Basic Concepts
Figure 12: Database Tables
Figure 13: Table Rows as Points in Space
Figure 14: Projection on the Plane
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Lesson: Understanding Motivation and Basic Concepts
Figure 15: Projection on an Axis
Components of Database Tables
Figure 16: Components Of Database Tables
Figure 17: Multi-Column Keys
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Unit 1: Motivation and Basic Concepts
Figure 18: Multi-Column Keys
Figure 19: Multiple Keys
Figure 20: Foreign-Keys
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Lesson: Understanding Motivation and Basic Concepts
Figure 21: Multi-Column Foreign-Keys
Figure 22: Multiple Foreign-Keys
Figure 23: Self-Referencing Foreign-Keys
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Unit 1: Motivation and Basic Concepts
LESSON SUMMARY
You should now be able to:
14
●
Understand the motivation for and foundation of the relational model
●
Understand SQL and its relation to the relation model
●
Understand database tables as the most important database objects
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Unit 1
Lesson 2
15
Understanding the Sample Data
LESSON OVERVIEW
This lessons explains the sample data models and data used throughout lessons and
exercises of this course.
LESSON OBJECTIVES
After completing this lesson, you will be able to:
●
Understand the sample database used throughout the course
The Sample Data Model
The examples used in the lessons and on the figures of this course are based on a fictional car
registration office.
Example: Registration Office
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●
The database of a fictional registration office will serve as the basis for further
explanations.
The tables in this database have been specifically tailored to the SQL course and are not an
example of good database design.
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The officials working in the fictional registration office have a manager.
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Each vehicle is registered for exactly one owner (or is unregistered).
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There is a list of vehicles that have been reported stolen.
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Owners, who have at least three vehicles registered, are assigned to one or multiple
contacts.
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Unit 1: Motivation and Basic Concepts
Figure 24: Registration Office Example
Table 2: The Officials Table
The following shows the content of the Official database table as an example.
For the content of the other sample tables check the handbook.
Official PNr
Name
Overtime
Salary
Manager
P01
Mr A
10
A09
P04
P02
Mr B
10
A10
P04
P03
Ms C
20
A09
P04
P04
Ms D
NULL
A12
P09
P05
Mr E
10
A08
P08
P06
Mr F
18
A09
P09
P07
Ms G
22
A11
P08
P08
Ms H
NULL
A13
P09
P09
Mr I
NULL
A14
NULL
Table 3: The Owner Table
Database table Owner contains the information about the car owners. Note that real persons
have a birthday, but companies don’t.
Owner
16
OwnerID
Name
Birthday
City
H01
Ms T
20.06.1934
Wiesloch
H02
Ms U
11.05.1966
Hockenheim
H03
SAP AG
NULL
Walldorf
H04
HDM AG
NULL
Heidelberg
H05
Mr V
21.04.1952
Leimen
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Lesson: Understanding the Sample Data
Owner
OwnerID
Name
Birthday
City
H06
Ms W
01.06.1957
Wiesloch
H07
IKEA
NULL
Walldorf
H08
Mr X
30.08.1986
Walldorf
H09
Ms Y
10.02.1986
Sinsheim
H10
Mr Z
03.02.1986
Ladenburg
Table 4: The Contact Table
Database table Contact relates a car owner owning more than two cars to the officials who are
the respective contact persons.
Contact
PersNumber
OwnerID
P01
H03
P01
H04
P01
H07
P04
H03
P04
H04
P08
H04
P08
H07
P09
H03
Figure 25: Cars
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Unit 1: Motivation and Basic Concepts
Figure 26: Stolen Cars
Figure 27: Registration Office and EU
Figure 28: Owner (EU-Wide)
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Lesson: Understanding the Sample Data
Figure 29: Cars (EU-Wide)
LESSON SUMMARY
You should now be able to:
●
Understand the sample database used throughout the course
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Unit 1: Motivation and Basic Concepts
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UNIT 2
Data From a Table Or View
Lesson 1
Using Data from a Table or View
Exercise 1: Using the SELECT Statement to Read from a Single Table
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UNIT OBJECTIVES
●
Write simple database queries using SQL‘s SELECT statement and project columns in and
out of queries using the SELECT clause
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Calculate column values, use built-in functions and the CASE clause in column lists
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Avoid duplicates in SELECT statement result sets
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Limit results sets to a given number of rows and browse through result sets
●
Ensure a specific order in result sets
●
Restrict the result set using the WHERE clause
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Unit 2
Lesson 1
22
Using Data from a Table or View
LESSON OVERVIEW
The lesson covers the foundation of how to retrieve data from a database.
LESSON OBJECTIVES
After completing this lesson, you will be able to:
●
Write simple database queries using SQL‘s SELECT statement and project columns in and
out of queries using the SELECT clause
●
Calculate column values, use built-in functions and the CASE clause in column lists
●
Avoid duplicates in SELECT statement result sets
●
Limit results sets to a given number of rows and browse through result sets
●
Ensure a specific order in result sets
●
Restrict the result set using the WHERE clause
Overview of the SELECT Statement
The SELECT statement is an SQL statement, which is used to read data from a database table
or view. The SELECT statement is the central construct for read access to data, of the Data
Manipulation Language.
A SELECT statement must contain a SELECT and a FROM clause and can contain a WHERE,
GROUP BY, HAVING and ORDER BY clause.
Figure 30: The SELECT Statement: Mandatory and Optional Clauses
The SELECT Clause in its Most Simple Form
The SELECT clause specifies the Projection List, in other words which columns to read from
the data or which columns to include in the result set of the SELECT statement.
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Lesson: Using Data from a Table or View
Figure 31: Multiple Columns in Projection List
Figure 32: Sequence of Columns in the Projection List
Figure 33: All Columns Asterisk in Projection List
The following options are also supported:
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You can include a single column in the projection list.
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You can include the same column in the projection list repeatedly.
●
You can combine using the asterisk and explicitly named columns.
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Unit 2: Data From a Table Or View
●
You can use the asterisk repeatedly in the projection list.
We do not recommend using the last three options because each makes the result set difficult
to understand and work with.
Naming Columns of the Result Set
The column names of the result set can differ from the column names in the table selected
from. In other words you can “rename” result columns.
Note:
Renaming columns in the SELECT clause has no effect on the table selected from.
It only affects the result set returned.
Figure 34: Renaming Columns in the Projection
Column names not enclosed in quotation marks are converted to upper case. A column name
has to be enclosed in quotation marks if one of the following applies:
●
●
The column name should not be converted to upper case but be kept exactly as spelled in
your statement.
The column name contains a space, a punctuation mark, a colon or any other special
character.
Note:
The rule also applies to the names of the original database tables and their
columns: The SQL parser converts names not enclosed in quotation marks to
upper case. It then performs a case-sensitive search for the corresponding
database objects.
The keyword AS is optional.
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Lesson: Using Data from a Table or View
Figure 35: Keyword AS is Optional
You can use column names of the table selected from to rename other columns of the result
set, e.g. to exchange column names.
Figure 36: Reusing Existing Column Names for Other Columns
The SELECT Clause: Calculated Columns
The SELECT clause can also contain columns the value of which is computed dynamically,
typically based on the values of other columns.
In the most simple form, these calculated column can be literal values. Both string and
numerical literals are supported.
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Unit 2: Data From a Table Or View
Figure 37: Literal Values in the SELECT Clause
Note:
String values are enclosed in single quotes. This can lead to confusion if column
names are enclosed in quotation marks as well.
Numerical values are not enclosed in any quotes.
Figure 38: Distinguish String and Numerical Literals for Column Names
The projection list can even comprise of one or more literal values only. Note that the result
set still contains as many records, all comprising of the same literal values only, as the table
select from contains. If your intent is to select a single “literal record”, you can use the built-in
table DUMMY.
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Lesson: Using Data from a Table or View
Figure 39: Built-In Table DUMMY
Figure 40: Calculated Columns
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Unit 2: Data From a Table Or View
Figure 41: NULL Values and Calculated Columns
Figure 42: Naming Calculated Columns
SAP HANA supports using functions to calculate columns of the projection list. Different
functions are available for different data types of the input columns and values.
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Lesson: Using Data from a Table or View
Figure 43: Functions
Figure 44: Function Calls can be Nested
Table 5: Functions supported by SAP HANA (Selection)
SAP HANA supports a rich list of built-in function. The following is only a selection.
Function
Explanation
ADD_YEAR(<date>, <number>)
Adds a <number> of years to the specified <date>
CURRENT_DATE
Determines the current date
ABS(<number>)
Determines the absolute value
ROUND(<number>)
Rounds the <number>
SQRT(<number>)
Determines the square root of <number>
UPPER(<string>)
Converts the <string> to upper case
SUBSTR(<string>,<start>,<len>)
Extracts a substring of <string>
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Unit 2: Data From a Table Or View
Function
Explanation
CONCAT(<string>,<string>)
Concatenates two strings. Equivalent to the binary operator ||.
COALESCE(<expressions>)
Returns the result of the first element of the list of <expressions> that does not evaluate to NULL.
See the SAP HANA SQL Reference available on the SAP Help Portal for the full list of
supported built-in functions. http://help.sap.com/hana_platform#section6
You can also create scalar User-Defined Functions and use the in the same was as built-in
functions.
The CASE Expression
A special way of calculating column values is using a CASE expression. This expression allows
you determine a column value based on conditions, in other words to include an IF-THENELSE type of logic in the projection list.
Figure 45: Case Expression: Basic Form
Figure 46: Case Expression: Result
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Lesson: Using Data from a Table or View
Figure 47: Case Expression: Case not Handled
Figure 48: Case Expression: Using Columns in the THEN and ELSE Branches
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Unit 2: Data From a Table Or View
Figure 49: Case Expression: Result
Eliminating Duplicates in Result Sets
An important difference between SQL and the Relational Database Model is that SQL is multiset oriented, which allows for duplicate entries in result sets. However, you won’t notice this
as long as the database table you select from have a primary key and once you select all of the
columns.
Figure 50: No Duplicates With Primary Key
If you project a table and don’t include the full primary key, duplicate entries can result.
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Lesson: Using Data from a Table or View
Figure 51: Duplicates By Projection
In most cases, this is not intended and practical, as there is no way to distinguish the
duplicate entries from each other in the result set. This is why duplicate entries can be
removed explicitly using keyword DISTINCT. Note that duplicates are not removed by default
to speed up data retrieval.
Figure 52: Duplicate Elimination using DISTINCT
Figure 53: Avoid Duplicate Elimination Using ALL
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Unit 2: Data From a Table Or View
Figure 54: Duplicate Elimination: Treatment of NULL Values
Figure 55: Duplicates With A Multi-Column Result Set
Figure 56: DISTINCT Eliminates Duplicates In Multi-Column Result Set
In principle, you could always include the DISTINCT keyword in SELECT statement to be sure
to never get duplicates, since DISTINCT is compatible with using an asterisk in the projection
list. Note, however, that using DISTINCT comes at a small performance penalty.
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Lesson: Using Data from a Table or View
Figure 57: Duplicate Elimination
The TOP N Clause
You can use the TOP N clause as part of the SELECT clause to specify that the result set
should contain at most N rows.
Figure 58: Limiting the Result Set to a Maximum Number of Rows
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Unit 2: Data From a Table Or View
Figure 59: TOP N Clause can be Combined with DISTINCT
Figure 60: TOP N Clause with Fewer Rows than N
Figure 61: TOP N Clause with N=0
The LIMIT and OFFSET Clauses
The TOP N clause allows you to limit the result set to a specified maximum number of rows,
but it does not allow you to retrieve the “next” N rows. To browse through a result in a pageby-page way, you can use the LIMIT and OFFSET clauses.
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Lesson: Using Data from a Table or View
Figure 62: The LIMIT Clause as Alternative to TOP N
Figure 63: Combine LIMIT and OFFSET to Browse through a Result Set
Note:
The LIMIT and OFFSET clauses are not part of the SELECT clause, but are the last
clauses of a SELECT statement.
The ORDER BY Clause
By default, the result returned by a query in SAP HANA has no defined and no reliable sort
order. Executing the same query repeatedly can result in different sort orders.
Note:
Query results in the SAP HANA database are not sorted by primary key by default.
There is no defined and no reliable default sort order.
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Unit 2: Data From a Table Or View
You can use the ORDER BY clause to explicitly specify the sort order of the result set and this
way get a reliable sort order.
Figure 64: The ORDER BY Clause
Figure 65: Descending Sort Order
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Lesson: Using Data from a Table or View
Figure 66: Ascending Sort Order
Figure 67: Sorting by a Column not in the Result
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Unit 2: Data From a Table Or View
Figure 68: Sorting by Multiple Columns
Figure 69: Referencing Columns by Position
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Lesson: Using Data from a Table or View
Figure 70: Sorting and Column Renaming
Figure 71: Sorting by Calculated Columns
You can also use functions in the expression in the ORDER BY clause.
The WHERE Clause
The WHERE clause is used to select a subset of rows from the data source based on a specified
condition. Only the rows matching the condition are returned.
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Unit 2: Data From a Table Or View
Figure 72: WHERE Clause
Figure 73: WHERE Clause
Condition expressions in SQL are very similar to condition expressions as known from
programming languages such as ABAP, Java and so on. The columns of the data source take
over the roles of variables used in these languages.
●
●
●
A condition can refer to string-like or to numeric columns. The column used in the
condition do not have to be included in the projection list.
Atomic conditions can be combined to complex and nested conditions using the usual
logical operators AND, OR and NOT and parentheses to override operator precedence rules.
A condition can make use of functions, including nested functions.
An atomic condition typically contains a comparison operator. The usual comparison
operators =, <>, <, <=, > and >= are supported. But SQL supports a few additional
comparison operators.
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Lesson: Using Data from a Table or View
Figure 74: Comparison Operator IS NULL
Figure 75: Comparison Operator IS NOT NULL
Figure 76: Comparison Operator IN
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Unit 2: Data From a Table Or View
Figure 77: Comparison Operator BETWEEN ... AND
Figure 78: Comparison Operator LIKE
Figure 79: Wildcards in the LIKE Predicate
44
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Lesson: Using Data from a Table or View
Figure 80: Escaping Wildcards in the LIKE Predicate
Figure 81: Escaping Escape Characters in the LIKE Predicate
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45
Unit 2: Data From a Table Or View
Figure 82: A Complex WHERE Clause
Figure 83: Operator Precedence
46
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Unit 2
Exercise 1
47
Using the SELECT Statement to Read from a
Single Table
Business Example
Your company created a few database tables using the SAP HANA platform to manage
rudimentary information about its employees and departments. The database tables have
been created as follows:
CREATE COLUMN TABLE EMPLOYEE
-- data about company's emplyees
( DNumber
CHAR(7),
-- personnel number
Name
VARCHAR(20)
NOT NULL, -- employee's name
DateOfBirth
DATE,
-- employee's date of birth
RemainderDays
INTEGER,
-- # of vacation days left
DepID
CHAR(3),
-- department employee works for
PRIMARY KEY(DNumber)
);
CREATE COLUMN TABLE Department -- the company's departments
( DepID
CHAR(3) PRIMARY KEY, -- department ID
Name
VARCHAR(20),
-- department name
Status
VARCHAR(10)
DEFAULT 'profitable'
);
You are asked to answer a set of questions about employees and departments using SQL
queries against the data.
Note that this example used throughout the exercises represents a simplistic model for
storing information about a company’s employees and organizational structure.
Log on to SAP HANA and Open SQL Console
Establish a connection to the SAP HANA Server to be prepared for writing and executing SQL
queries.
1. Open SAP HANA Studio.
2. Add a new system. Use the server name, user ID and password provided by your
instructor.
3. Open an SQL Console connected to the SAP HANA system.
Using the SELECT Statement to Read from a Single Table
Write and execute a SELECT statement for each of the following tasks to answer the question
posed.
1. Create a list of employees, including all columns.
2. Write a query to determine which employees have fewer than 30 remaining vacation days.
3. Create a list of the department IDs the different employees work for. The list should not
contain duplicates.
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47
Unit 2: Data From a Table Or View
4. Write a query to determine which employees have fewer than 30 remaining vacation days.
Sort the output descending by RemainderDays and - for employees with the same number
of RemainderDays - ascending by DNumber.
5. Write a query to determine the employees for which the remaining vacation days are
unknown.
6. Write a query to determine the employees for which the remaining vacation days known.
7. Write a query to determine how many remaining vacation days can be found for the
different employees in the company. Each quantity number should be displayed only once.
8. Create a list of employees. Only columns DNumber and Name are relevant.
9. Create a list of employees. Only columns DNumber and Name are relevant. The output
column DNumber should be renamed to “PERSNUMBER”.
10. Create a list of employees. Only columns DNumber and Name are relevant. The output
column DNumber should be renamed to “Personal Number” (with small letters).
11. Write a query to determine how many remaining vacation days “Ms D” has.
12. Write a query to determine which employees have exactly 10 remaining vacation days.
13. The value of a vacation day is estimated at EUR 300.- (all-in). Create a list of employees
including the calculated value of the remaining vacation days.
14. The value of a vacation day is estimated at EUR 300.- (all-in). Write a query to determine
which employees have remaining vacation days with a value of at least EUR 600.15. Write a query to determine which numbers of remaining vacation days can be found in the
company. Each quantity number should be displayed only once.
16. Write a query to determine which employees work for department “A04” and have exactly
20 remaining vacation days.
17. Write a query to determine which employees work in department “A01” or “A04”. Use OR
18. Write a query to determine which employees work in department "A01" or "A04". Use IN
19. Write a query to determine which employees not working for department “A01” have
exactly 10 remaining vacation days.
20. Write a query to determine which employee names include an “L”.
21. Write a query to determine which employees have a name beginning with “M” and in the
fourth place a “K”.
48
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Unit 2
Solution 1
49
Using the SELECT Statement to Read from a
Single Table
Business Example
Your company created a few database tables using the SAP HANA platform to manage
rudimentary information about its employees and departments. The database tables have
been created as follows:
CREATE COLUMN TABLE EMPLOYEE
-- data about company's emplyees
( DNumber
CHAR(7),
-- personnel number
Name
VARCHAR(20)
NOT NULL, -- employee's name
DateOfBirth
DATE,
-- employee's date of birth
RemainderDays
INTEGER,
-- # of vacation days left
DepID
CHAR(3),
-- department employee works for
PRIMARY KEY(DNumber)
);
CREATE COLUMN TABLE Department -- the company's departments
( DepID
CHAR(3) PRIMARY KEY, -- department ID
Name
VARCHAR(20),
-- department name
Status
VARCHAR(10)
DEFAULT 'profitable'
);
You are asked to answer a set of questions about employees and departments using SQL
queries against the data.
Note that this example used throughout the exercises represents a simplistic model for
storing information about a company’s employees and organizational structure.
Log on to SAP HANA and Open SQL Console
Establish a connection to the SAP HANA Server to be prepared for writing and executing SQL
queries.
1. Open SAP HANA Studio.
a) Choose Start → All Programs → SAP HANA → SAP HANA Studio from the Windows
Start Menu.
b) In case the Welcome page is open, close the Welcome page or press the Workbench
link.
2. Add a new system. Use the server name, user ID and password provided by your
instructor.
a) Make sure you are in the SAP HANA Modeler perspective. If not, change perspective
using the buttons in the upper right-hand corner.
b) Right-click anywhere in the Systems window on the left and choose Add System....
c) Enter host name and instance number provided by your instructor and a description,
then press button Next >.
d) Enter user ID and password provided by your instructor, then choose Finish.
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49
Unit 2: Data From a Table Or View
3. Open an SQL Console connected to the SAP HANA system.
a) In the Systems frame, select the SAP HANA System or the Catalog folder underneath
it.
b) Choose context menu item Open SQL Console or press the corresponding button in
the toolbar of the Systems frame.
Using the SELECT Statement to Read from a Single Table
Write and execute a SELECT statement for each of the following tasks to answer the question
posed.
1. Create a list of employees, including all columns.
a) SELECT *
FROM Employee;
2. Write a query to determine which employees have fewer than 30 remaining vacation days.
a) SELECT *
FROM Employee
WHERE RemainderDays < 30;
3. Create a list of the department IDs the different employees work for. The list should not
contain duplicates.
a) SELECT DISTINCT DEPID
FROM Employee;
4. Write a query to determine which employees have fewer than 30 remaining vacation days.
Sort the output descending by RemainderDays and - for employees with the same number
of RemainderDays - ascending by DNumber.
a) SELECT * FROM EmployeeWHERE RemainderDays < 30 ORDER BY
RemainderDays DESC, DNumber ASC;
5. Write a query to determine the employees for which the remaining vacation days are
unknown.
a) SELECT *
FROM Employee
WHERE RemainderDays IS NULL;
6. Write a query to determine the employees for which the remaining vacation days known.
a) SELECT *
FROM Employee
WHERE RemainderDays IS NOT NULL;
7. Write a query to determine how many remaining vacation days can be found for the
different employees in the company. Each quantity number should be displayed only once.
a) SELECT DISTINCT RemainderDays
RemainderDays IS NOT NULL;
FROM Employee
WHERE
8. Create a list of employees. Only columns DNumber and Name are relevant.
a) SELECT DNumber, Name
FROM Employee;
9. Create a list of employees. Only columns DNumber and Name are relevant. The output
column DNumber should be renamed to “PERSNUMBER”.
a) SELECT DNumber AS PersNumber, Name
FROM Employee;
10. Create a list of employees. Only columns DNumber and Name are relevant. The output
column DNumber should be renamed to “Personal Number” (with small letters).
a) SELECT DNumber AS "Personal Number", Name
FROM Employee;
11. Write a query to determine how many remaining vacation days “Ms D” has.
a) SELECT RemainderDays
FROM Employee
WHERE Name = 'Ms D';
12. Write a query to determine which employees have exactly 10 remaining vacation days.
50
© Copyright. All rights reserved.
Lesson: Using Data from a Table or View
a) SELECT *
FROM Employee
WHERE RemainderDays = 10;
13. The value of a vacation day is estimated at EUR 300.- (all-in). Create a list of employees
including the calculated value of the remaining vacation days.
a) SELECT *, RemainderDays * 300 AS RemainderDaysValue
Employee;
FROM
14. The value of a vacation day is estimated at EUR 300.- (all-in). Write a query to determine
which employees have remaining vacation days with a value of at least EUR 600.a) SELECT *, RemainderDays * 300 AS RemainderDayValue
Employee WHERE RemainderDays * 300 >= 600;
FROM
15. Write a query to determine which numbers of remaining vacation days can be found in the
company. Each quantity number should be displayed only once.
a) SELECT DISTINCT RemainderDays
RemainderDays IS NOT NULL;
FROM Employee
WHERE
16. Write a query to determine which employees work for department “A04” and have exactly
20 remaining vacation days.
a) SELECT *
= 20;
FROM Employee
WHERE DepID = 'A04' AND RemainderDays
17. Write a query to determine which employees work in department “A01” or “A04”. Use OR
a) SELECT *
FROM Employee
WHERE DepID = 'A01' OR DepID = 'A04';
18. Write a query to determine which employees work in department "A01" or "A04". Use IN
a) SELECT *
FROM Employee
WHERE DepID IN ('A01', 'A04');
19. Write a query to determine which employees not working for department “A01” have
exactly 10 remaining vacation days.
a) SELECT *
= 10;
FROM Employee
WHERE DepID <> 'A01' AND RemainderDays
20. Write a query to determine which employee names include an “L”.
a) SELECT *
FROM Employee
WHERE Name LIKE '%L%';
21. Write a query to determine which employees have a name beginning with “M” and in the
fourth place a “K”.
a) SELECT *
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FROM Employee
WHERE Name LIKE 'M__K%';
51
Unit 2: Data From a Table Or View
LESSON SUMMARY
You should now be able to:
●
52
Write simple database queries using SQL‘s SELECT statement and project columns in and
out of queries using the SELECT clause
●
Calculate column values, use built-in functions and the CASE clause in column lists
●
Avoid duplicates in SELECT statement result sets
●
Limit results sets to a given number of rows and browse through result sets
●
Ensure a specific order in result sets
●
Restrict the result set using the WHERE clause
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UNIT 3
Aggregation of Data
Lesson 1
Aggregating Data
Exercise 2: Use Aggregate Expressions
54
65
UNIT OBJECTIVES
●
List the most important aggregate functions supported by HANA and use them to
determine aggregated values on table columns using a single SELECT statement
●
Determine aggregated values for groups of rows, using the GROUP BY clause
●
Filter groups using the HAVING clause
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53
Unit 3
Lesson 1
54
Aggregating Data
LESSON OVERVIEW
This lesson covers how you can perform certain calculations on a database table column
across multiple rows of the database table.
LESSON OBJECTIVES
After completing this lesson, you will be able to:
●
List the most important aggregate functions supported by HANA and use them to
determine aggregated values on table columns using a single SELECT statement
●
Determine aggregated values for groups of rows, using the GROUP BY clause
●
Filter groups using the HAVING clause
Calculations Across Multiple Rows: Aggregation
Function expressions allow calculating the values of columns based on the values of other
columns on a row-by-row basis. But in many cases, you may want to calculate a single value
across the values of multiple rows in one or more columns. You can use aggregate
expressions for this purpose.
Figure 84: COUNT(*) Expression
54
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Lesson: Aggregating Data
Figure 85: Renaming the Result of an Aggregate Expression
Figure 86: COUNT(<column>) Expressions
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55
Unit 3: Aggregation of Data
Figure 87: COUNT(DISTINCT <column>) Expression
Figure 88: Differences between the COUNT Expressions
Figure 89: MIN and MAX Expressions
56
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Lesson: Aggregating Data
Figure 90: Combining Aggregate and Function Expressions
Note that, in general, the order in which functions and aggregate expressions are nested
matters. As an example, the following two SELECT statements lead to different results:
SELECT ABS( MAX(0-HP) ) FROM Car; -- Result: 75
SELECT MAX( ABS(0-HP) ) FROM Car; -- Result: 260
Figure 91: Aggregate Expressions
Figure 92: SUM(<column>) and AVG(<column>) Expressions
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57
Unit 3: Aggregation of Data
Figure 93: SUM(DISTINCT <column>) and AVG(DISTINCT <column>) Expressions
Table 6: Aggregate Expressions Supported by SAP HANA
The following table lists a selection of the aggregate expressions currently supported by SAP
HANA
Aggregate Name
Description
COUNT
Returns the number of rows.
MIN, MAX
Returns the minimum, respectively maximum value of
an input column with a numeric data type.
SUM, AVG
Returns the sum, respectively the arithmetic mean value
of an input column with a numeric data type.
MEDIAN
Finds the statistical median of an input column with a
numeric data type.
VAR, STDDEV
Returns the variance, respectively the standard deviation. STDDEV is the square root of VAR.
CORR, CORR_SPEARMAN
Computes the Pearson product momentum correlation
coefficient, respectively the Spearman's rank correlation
coefficient between two columns.
STRING_AGG
Returns the concatenated string.
The GROUP BY Clause
The examples so far all calculated a single value across all rows of a table that satisfied the
WHERE condition of the SELECT statement. Sometimes you may need such aggregations not
for all rows in total, but to produce different values for different subsets. This is here grouping
and the GROUP BY clause come into play.
58
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Lesson: Aggregating Data
Figure 94: COUNT(*) with GROUP BY Clause
Figure 95: MAX with GROUP BY Clause
Figure 96: Treatment of NULL Values
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59
Unit 3: Aggregation of Data
Figure 97: Grouping and Sorting
Figure 98: Combining Functions and Grouping
Figure 99: Grouping and WHERE Clause
60
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Lesson: Aggregating Data
Figure 100: Grouping by Several Columns
Note:
In SAP HANA SQL, you have to explicitly include each column in the group by
clause that is not used in an aggregate expression.
Figure 101: Grouping and Renaming Columns
The HAVING Clause
When using grouping, you can discard some of the resulting groups in a way similar to how
you exclude rows from a non-aggregated result set. The corresponding keyword is HAVING
and behaves similar to the WHERE clause, but affects the resulting groups instead of which
rows are considered when calculating the aggregate values.
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61
Unit 3: Aggregation of Data
Figure 102: HAVING
Figure 103: HAVING
62
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Lesson: Aggregating Data
Figure 104: SELECT statement
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63
Unit 3: Aggregation of Data
64
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Unit 3
Exercise 2
65
Use Aggregate Expressions
1. What is the number of employees?
2. How many employees have exactly 20 remaining vacation days?
3. In how many different departments do the employees work?
4. What is the maximum number of remaining vacation days?
5. What is the average number of remaining vacation days? Employees whose remaining
vacation days are unknown should not be taken into account.
6. How many remaining vacation days have the employees accumulated in total?
7. The minimum, average, maximum, and total number of remaining vacation days should be
displayed with a single query.
8. What is the difference between the maximum remaining vacation days and the average of
remaining vacation days?
9. The total number of remaining vacation days should be displayed for each department.
10. The total number of remaining vacation days should be displayed for each department.
Only departments with more than 30 days should be taken into account. The display of
the total number of days should be sorted descending.
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65
Unit 3
Solution 2
66
Use Aggregate Expressions
1. What is the number of employees?
a) SELECT COUNT(*)
FROM Employee;
2. How many employees have exactly 20 remaining vacation days?
a) SELECT COUNT(*)
FROM Employee
WHERE RemainderDays = 20;
3. In how many different departments do the employees work?
a) SELECT COUNT(DISTINCT DepID)
FROM Employee;
4. What is the maximum number of remaining vacation days?
a) SELECT MAX(RemainderDays)
FROM Employee;
5. What is the average number of remaining vacation days? Employees whose remaining
vacation days are unknown should not be taken into account.
a) SELECT AVG(RemainderDays)
FROM Employee;
6. How many remaining vacation days have the employees accumulated in total?
a) SELECT SUM(RemainderDays)
FROM Employee;
7. The minimum, average, maximum, and total number of remaining vacation days should be
displayed with a single query.
a) SELECT MIN(RemainderDays) AS "min", AVG(RemainderDays) AS "avg",
MAX(RemainderDays) AS "max",
SUM(RemainderDays) AS "sum"
FROM Employee;
8. What is the difference between the maximum remaining vacation days and the average of
remaining vacation days?
a) SELECT MAX(RemainderDays) - AVG(RemainderDays)
FROM Employee;
9. The total number of remaining vacation days should be displayed for each department.
a) SELECT DepID, SUM(RemainderDays) FROM Employee GROUP BY DepID;
10. The total number of remaining vacation days should be displayed for each department.
Only departments with more than 30 days should be taken into account. The display of
the total number of days should be sorted descending.
a) SELECT DepID, SUM(RemainderDays) FROM Employee GROUP BY DepID
HAVING SUM(RemainderDays) > 30 ORDER BY 2 DESC;
66
© Copyright. All rights reserved.
Lesson: Aggregating Data
LESSON SUMMARY
You should now be able to:
●
List the most important aggregate functions supported by HANA and use them to
determine aggregated values on table columns using a single SELECT statement
●
Determine aggregated values for groups of rows, using the GROUP BY clause
●
Filter groups using the HAVING clause
© Copyright. All rights reserved.
67
Unit 3: Aggregation of Data
68
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UNIT 4
Data From Multiple Tables Part
I
Lesson 1
Reading Data from Multiple Tables- Part 1
Exercise 3: Use JOIN to Combine Data from Several Tables
70
91
UNIT OBJECTIVES
●
●
Combine the result of several select statements using the UNION, INTERSECT and
EXCEPT statements
List the various types of JOIN constructs and use the appropriate JOIN construct to
combine data from several tables using a single query
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69
Unit 4
Lesson 1
70
Reading Data from Multiple Tables- Part 1
LESSON OVERVIEW
This lesson covers how data can be retrieved and combined from several tables in a single
statement, using UNIONs and JOINs.
LESSON OBJECTIVES
After completing this lesson, you will be able to:
●
●
Combine the result of several select statements using the UNION, INTERSECT and
EXCEPT statements
List the various types of JOIN constructs and use the appropriate JOIN construct to
combine data from several tables using a single query
Reading from Multiple Tables: Overview
The SELECT statement is not limited to reading from a single table. You can use it to read
from several tables using a single statement. The following lists the options that are available,
depending on the exact needs.
●
You can combine the results of several partial queries to form the union of the results,
determine their intersection or difference:
-
UNION ALL and UNION merge the results of partial queries to form their union.
-
INTERSECT is used to determine the elements contained in all partial query results.
-
●
You can use JOIN to combine data from several tables in common rows of a result set.
-
-
●
CROSS JOIN, INNER JOIN and OUTER JOIN are the three types of JOINs. The are
explained in the following.
The two syntax options implicit JOIN and explicit JOIN are available.
You can use nested queries (sub queries) to embed queries for information lookup in other
queries.
-
70
EXCEPT is used to determine the elements contained in one partial result, but not the
second partial result.
An uncorrelated sub query should be distinguished and favored over a correlated sub
query.
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Lesson: Reading Data from Multiple Tables- Part 1
UNION and UNION ALL
Figure 105: UNION [ALL]
Figure 106: Result Header is Determined by the First SELECT Clause
Figure 107: Renaming Columns with UNION [ALL]
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71
Unit 4: Data From Multiple Tables Part I
Figure 108: UNION ALL can Lead to Duplicates
Figure 109: Avoiding Duplicates using UNION
72
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Lesson: Reading Data from Multiple Tables- Part 1
Figure 110: UNION [ALL] can Handle more than Two SELECTs
INTERSECT and EXCEPT
Figure 111: INTERSECT [DISTINCT]
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73
Unit 4: Data From Multiple Tables Part I
Figure 112: EXCEPT [DISTINCT]
Cross Join
You can use a cross join, also called Cartesian product, to create a result set in which each
row of each partial query involved is combined with each row of the other partial queries
involved.
Figure 113: CROSS JOIN: Explanation
74
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Lesson: Reading Data from Multiple Tables- Part 1
Figure 114: CROSS JOIN with Implicit Syntax
Figure 115: CROSS JOIN with Explicit Syntax
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75
Unit 4: Data From Multiple Tables Part I
Figure 116: CROSS JOIN and WHERE Clause
Figure 117: CROSS JOIN
You can also cross-join a table with itself.
In general, note that a cross join can quickly lead to a very large result set, as its cardinality
(number or rows) equals the product of the cardinalities of the queries it is comprised of. It
can be easy to produce a cross join by mistake, because its syntax looks very similar to that of
the inner join.
Table Aliases
Different tables can have columns with the same name. This can lead to ambiguities when
joining such tables, for example when a column name that appears in two or more of the
tables involved is used in the projection list, the WHERE condition or in a join condition. You
have to resolve such ambiguities by qualifying the non-unique column name, that is by
specifying the table the column belongs to. You can do this using the full table name as
qualifier.
76
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Lesson: Reading Data from Multiple Tables- Part 1
Figure 118: Qualifying a Column Name using the Table Name
Having to repeat the full table name as qualifier each time can be cumbersome. This is why
you can use table aliases to introduce abbreviated table names and use the abbreviated form
as qualifier in the same SELECT statement.
Figure 119: Table Aliases
Figure 120: Qualified Names can be Mixed with Unqualified Names
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77
Unit 4: Data From Multiple Tables Part I
Figure 121: Supported Combinations of Qualified and Unqualified Column Names
Inner Join
An inner join is used to combine information from one table with corresponding information
from another table, and to only include such rows in the result for which corresponding
information is available.
Figure 122: Inner Join: Explanation
78
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Lesson: Reading Data from Multiple Tables- Part 1
Figure 123: Inner Join — Implicit Syntax
Figure 124: Inner Join — Explicit Syntax
Note:
The following examples will show the explicit syntax option only. Using the implicit
syntax is more error-prone since the join condition cannot be distinguished from
additional parts of a WHERE condition. This makes it easy to forget the join
condition and produce a cross join by mistake.
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79
Unit 4: Data From Multiple Tables Part I
Figure 125: Keyword INNER is Optional
Figure 126: Joins can be Combined with Projections
80
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Lesson: Reading Data from Multiple Tables- Part 1
Figure 127: Table Aliases can be Handy for Inner Joins
Figure 128: Projections can Reduce to One of the Tables Involved
Figure 129: Renaming Columns of the Result Works as Usual
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81
Unit 4: Data From Multiple Tables Part I
Figure 130: Aggregating, Grouping etc. also Work as Usual
Figure 131: The JOIN Condition can be Complex
Figure 132: Implicit INNER JOIN
82
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Lesson: Reading Data from Multiple Tables- Part 1
Figure 133: Combining Two Tables
Figure 134: Use of Different Comparison Operators
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83
Unit 4: Data From Multiple Tables Part I
Figure 135: Calculations and Functions in the JOIN Condition
Left Outer Join
The result set of an inner join does not contain the rows of any of the involved tables for which
no corresponding row can be found in the other tables involved according to the join
condition. Sometimes you may want to include such rows in the result set, to get the full set of
rows of one table and augment the data with data from other tables as far as available. In
addition to the inner join, three types of outer joins exist for this purpose:
●
●
●
Left outer join: includes all entries of the first (left) table even if no corresponding entries
are found in the second (right) table.
Right outer join: includes all entries of the second table even if no corresponding entries
are found in the first table.
Full outer join: is the union of left and right outer join.
In all three cases, the following applies:
●
●
84
A NULL value results for each column and row for which no corresponding data can be
retrieved according to the join condition.
SAP HANA only supports the explicit syntax option.
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Lesson: Reading Data from Multiple Tables- Part 1
Figure 136: LEFT OUTER Join: Explanation
Figure 137: LEFT OUTER Join: Example
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85
Unit 4: Data From Multiple Tables Part I
Figure 138: LEFT OUTER Join can be Combined with WHERE
Right Outer Join
Figure 139: RIGHT OUTER Join: Explanation
86
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Lesson: Reading Data from Multiple Tables- Part 1
Figure 140: RIGHT OUTER Join: Example
Figure 141: RIGHT OUTER Join can be Combined with WHERE
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87
Unit 4: Data From Multiple Tables Part I
Full Outer Join
Figure 142: FULL OUTER Join: Explanation
Figure 143: FULL OUTER Join: Example
88
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Lesson: Reading Data from Multiple Tables- Part 1
Figure 144: FULL OUTER Join can be Combined with WHERE
Each type of outer join supports joining a table with itself. The following shows the example of
a left outer join.
Figure 145: LEFT OUTER JOIN
Replacing keyword LEFT with RIGHT or FULL in the example works and leads to different
result sets.
●
●
Using RIGHT OUTER JOIN would lead to a result including each official name in the
MANAGER column and with only the names of the official who have a manager assigned in
the EMPLOYEE column.
USING FULL OUTER JOIN would lead to a result including each official name in the
MANAGER column and with each official name in the EMPLOYEE column.
Similar to what is supported for CROSS JOINs and INNER JOINs, you can also do the following
with OUTER JOINs:
●
Use any comparison operator in the join condition
●
Join more than two tables, and join a table with itself.
© Copyright. All rights reserved.
89
Unit 4: Data From Multiple Tables Part I
●
●
Rename columns
●
Eliminate duplicates using distinct
●
Order the result set, use aggregate expressions, a GROUP BY clause, a HAVING clause
●
90
Project the result set to a subset of columns. You do not have to include a column of every
table involved in the projection list.
Use table aliases and reference them in the projection list and join condition. You have to
qualify column names with the table names or use aliases if the column names are not
unique.
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Unit 4
Exercise 3
91
Use JOIN to Combine Data from Several Tables
1. Which employees work for department "A01" or A04"? Use UNION.
2. Which employees work for which department? Use an implicit JOIN without table alias.
Employees without department and departments without employees should not be taken
into account.
3. Which employees work for which department? Use an implicit JOIN with table aliases.
Employees without department and departments without employees should not be taken
into account.
4. Which employees work for which department? Use an explicit JOIN without table aliases.
Employees without department and departments without employees should not be taken
into account.
5. Which employees work for which department? Use an explicit JOIN with table aliases.
Employees without department and departments without employees should not be taken
into account.
6. Which employees work for which department? Employees without department should be
included, whereas departments without employees should be ignored. Use a LEFT OUTER
JOIN with table aliases.
7. Which employees work for which department? Employees without department should be
ignored, whereas departments without employees should be included. Use a RIGHT
OUTER JOIN with table aliases.
8. Which employees work for which department? Employees without department and
departments without employees should be included. Use a FULL OUTER JOIN with table
aliases.
9. Which employees work for the Sales department? Use an implicit Join with table aliases.
10. Which employees work for the Sales department? Use an explicit Join with table aliases.
11. Which department has accumulated how many remaining vacation days? Only colleagues
born after 03.02.1972 should be taken into account. The output of the number of days
should be sorted descending. Only departments that accumulated more than 30
remaining vacation days should be included.
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91
Unit 4
Solution 3
92
Use JOIN to Combine Data from Several Tables
1. Which employees work for department "A01" or A04"? Use UNION.
a) SELECT *
SELECT *
FROM Employee
FROM Employee
WHERE DepID = 'A01' UNION
WHERE DepID = 'A04';
2. Which employees work for which department? Use an implicit JOIN without table alias.
Employees without department and departments without employees should not be taken
into account.
a) SELECT Employee.Name, Department.Name
FROM Employee,
Department WHERE Employee.DepID = Department.DepID;
3. Which employees work for which department? Use an implicit JOIN with table aliases.
Employees without department and departments without employees should not be taken
into account.
a) SELECT m.Name, a.Name
m.DepID = a.DepID;
FROM Employee m, Department a
WHERE
4. Which employees work for which department? Use an explicit JOIN without table aliases.
Employees without department and departments without employees should not be taken
into account.
a) SELECT Employee.Name, Department.Name
FROM Employee JOIN
Department ON Employee.DepID = Department.DepID;
5. Which employees work for which department? Use an explicit JOIN with table aliases.
Employees without department and departments without employees should not be taken
into account.
a) SELECT m.Name, a.Name
m.DepID = a.DepID;
FROM Employee m JOIN Department a ON
6. Which employees work for which department? Employees without department should be
included, whereas departments without employees should be ignored. Use a LEFT OUTER
JOIN with table aliases.
a) SELECT m.Name, a.Name
FROM Employee m LEFT OUTER JOIN
Department a ON m.DepID = a.DepID;
7. Which employees work for which department? Employees without department should be
ignored, whereas departments without employees should be included. Use a RIGHT
OUTER JOIN with table aliases.
a) SELECT m.Name, a.Name
FROM Employee m RIGHT OUTER JOIN
Department a ON m.DepID = a.DepID;
8. Which employees work for which department? Employees without department and
departments without employees should be included. Use a FULL OUTER JOIN with table
aliases.
92
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Lesson: Reading Data from Multiple Tables- Part 1
a) SELECT m.Name, a.Name
FROM Employee m FULL OUTER JOIN
Department a ON m.DepID = a.DepID;
9. Which employees work for the Sales department? Use an implicit Join with table aliases.
a) SELECT m.*
FROM Employee m, Department a
a.DepID AND a.Name = 'Sales';
WHERE m.DepID =
10. Which employees work for the Sales department? Use an explicit Join with table aliases.
a) SELECT m.* FROM Employee m JOIN Department a ON m.DepID = a.DepID WHERE
a.Name = 'Sales';
11. Which department has accumulated how many remaining vacation days? Only colleagues
born after 03.02.1972 should be taken into account. The output of the number of days
should be sorted descending. Only departments that accumulated more than 30
remaining vacation days should be included.
a) SELECT a.Name, SUM(m.RemainderDays) AS DepartmentsVacation
FROM Employee m JOIN Department a ON m.DepID = a.DepID
WHERE
m.DateOfBirth > '1972-02-03' GROUP BY a.Name, a.DepID
HAVING
SUM(m.RemainderDays) > 30 ORDER BY DepartmentsVacation DESC;
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93
Unit 4: Data From Multiple Tables Part I
LESSON SUMMARY
You should now be able to:
●
●
94
Combine the result of several select statements using the UNION, INTERSECT and
EXCEPT statements
List the various types of JOIN constructs and use the appropriate JOIN construct to
combine data from several tables using a single query
© Copyright. All rights reserved.
UNIT 5
Data From Multiple Tables Part
II
Lesson 1
Reading Data from Multiple Tables- Part 2
Exercise 4: Use Nested Queries
96
107
UNIT OBJECTIVES
●
Explain the difference between uncorrelated and correlated sub-queries and use both to
query data from multiple tables in a single select statement
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95
Unit 5
Lesson 1
96
Reading Data from Multiple Tables- Part 2
LESSON OVERVIEW
This lesson covers how sub-queries can be used to retrieve data from several tables in a
single statement.
LESSON OBJECTIVES
After completing this lesson, you will be able to:
●
Explain the difference between uncorrelated and correlated sub-queries and use both to
query data from multiple tables in a single select statement
Sub Queries — Overview
●
●
In addition to UNION and JOIN, sub queries or nested queries also provide the option of
reading from multiple tables and views.
Nested Queries contain a so called “Sub Query”.
Figure 146: Sub Query: Explanation
A sub query can be useful to achieve the following:
●
●
●
96
To make a SELECT statement more “readable”.
To improve the performance of a SELECT statement – however, with a “perfect”
optimizer, this should not be the case.
To formulate a SELECT statement which cannot be formulated (or only extremely
cumbersome) without a sub query - which can be particularly the case for aggregate
expressions.
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Lesson: Reading Data from Multiple Tables- Part 2
Figure 147: Uncorrelated Versus Correlated Sub Query
Uncorrelated Sub Queries
Figure 148: Frequent Case: Combining IN with a Sub Query
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97
Unit 5: Data From Multiple Tables Part II
Figure 149: The ANY Expression
Figure 150: Uses of the ANY Expression
Figure 151: Example for > ANY
98
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Lesson: Reading Data from Multiple Tables- Part 2
Figure 152: Special Case: Single-Row Sub Query
Figure 153: The ALL Expression
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99
Unit 5: Data From Multiple Tables Part II
Figure 154: <= ALL Example
Figure 155: Special Case: Single-Row Sub Query
100
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Lesson: Reading Data from Multiple Tables- Part 2
Figure 156: Sub Queries can use Functions
Figure 157: Sub Query with a Multi-Column Result Set
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101
Unit 5: Data From Multiple Tables Part II
Figure 158: Sub Queries can be Nested
Figure 159: Sub Queries can be Combined with JOIN
Figure 160: Sub Queries are not Limited to the WHERE Clause
102
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Lesson: Reading Data from Multiple Tables- Part 2
Figure 161: Sub Query in the SELECT Clause
Correlated Sub Queries
Figure 162: Correlated Sub Query with EXISTS
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103
Unit 5: Data From Multiple Tables Part II
Figure 163: Table Aliases can be Optional
Figure 164: Correlated Sub Query with NOT EXISTS
Figure 165: Correlated Sub Query using Same Table
104
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Lesson: Reading Data from Multiple Tables- Part 2
Figure 166: Correlated Sub Query with >= ALL
Figure 167: Correlated Sub Queries can be Nested
Figure 168: Correlated Sub Query in the SELECT Clause
Using nested queries you can also perform the following:
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105
Unit 5: Data From Multiple Tables Part II
106
●
Restrict the projection list to certain columns
●
Explicitly rename the result columns
●
Sort the query result
●
Use grouping
●
Include aggregate expressions
●
Eliminate duplicates using DISTINCT
●
Involve the same table multiple times
●
Involve more than two tables
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Unit 5
Exercise 4
107
Use Nested Queries
1. Which employees work for the sales department? Use an uncorrelated sub query.
2. Which employees work for the sales department? Use a correlated sub query.
3. Which employees accumulated the most remaining vacation days? Use an uncorrelated
sub query with "MAX".
4. Which employees accumulated the most remaining vacation days? Use an uncorrelated
sub query without "MAX".
5. Which employees have above average remaining vacation days?
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107
Unit 5
Solution 4
108
Use Nested Queries
1. Which employees work for the sales department? Use an uncorrelated sub query.
a) SELECT *
FROM Employee
WHERE DepID = ANY (SELECT DepID
Department WHERE Name = 'Sales');
FROM
2. Which employees work for the sales department? Use a correlated sub query.
a) SELECT m.*
FROM Employee m WHERE EXISTS (SELECT * FROM
Department a WHERE a.DepID = m.DepID AND a.Name = 'Sales');
3. Which employees accumulated the most remaining vacation days? Use an uncorrelated
sub query with "MAX".
a) SELECT *
FROM Employee WHERE RemainderDays = (SELECT
MAX(RemainderDays) FROM Employee);
4. Which employees accumulated the most remaining vacation days? Use an uncorrelated
sub query without "MAX".
a) SELECT *
FROM Employee WHERE RemainderDays >= ALL (SELECT
RemainderDays FROM Employee WHERE RemainderDays IS NOT NULL);
5. Which employees have above average remaining vacation days?
a) SELECT *
FROM Employee WHERE RemainderDays > (SELECT
AVG(RemainderDays) FROM Employee);
108
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Lesson: Reading Data from Multiple Tables- Part 2
LESSON SUMMARY
You should now be able to:
●
Explain the difference between uncorrelated and correlated sub-queries and use both to
query data from multiple tables in a single select statement
© Copyright. All rights reserved.
109
Unit 5: Data From Multiple Tables Part II
110
© Copyright. All rights reserved.
UNIT 6
NULL Values
Lesson 1
Understanding NULL Values
Exercise 5: Understand NULL Values
112
117
UNIT OBJECTIVES
●
Interpret NULL values in databases and understand why their presence can lead to
unexpected query results
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111
Unit 6
Lesson 1
112
Understanding NULL Values
LESSON OVERVIEW
This lesson explains what NULL values are and why dealing with them correctly is important.
LESSON OBJECTIVES
After completing this lesson, you will be able to:
●
Interpret NULL values in databases and understand why their presence can lead to
unexpected query results
NULL Values: Interpretation and Origin
Interpretation of NULL Values
●
A NULL value signals that the corresponding value is unknown or does not exist.
●
NULL is a special value reserved by the DBMS.
●
-
NULL does not equal 0 for numeric data types.
-
NULL does not equal an empty string or a space character for string based data types.
A NULL value can express the following:
-
-
The corresponding value exists in principle, but is unknown temporarily or permanently
for the row at hand (such as a birthday of a person).
The corresponding value does not exist, for example the owner of an unregistered car.
NULL Values can Result in the Following Cases:
●
●
When reading rows (SELECT statement)
-
If the table queried from already contains NULL values
-
When using outer joins
When inserting or updating rows (INSERT and UPDATE statements)
-
-
●
If a NULL value is explicitly specified for a column and NULL values are allowed
When adding a column to a table (ALTER TABLE statement)
-
112
If no value for a column is provided, NULL values are allowed and no default value is
defined
If the table has at least one row, NULL values are allowed for the added column and no
default value is defined for the added column
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Lesson: Understanding NULL Values
Figure 169: NULL Values Lead to Trivalent Logic
Figure 170: Trivalent Logic Extends Logical Operators. Example: AND
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113
Unit 6: NULL Values
Figure 171: What Makes Trivalent Logic Difficult?
The introduction of the logical value unknown and the way how expressions are evaluated by
first evaluating their atomic sub expressions leads to an unexpected result: Logical
expressions that always evaluate to true in usual programming languages and according to
common sense no longer evaluate to true in all cases. The following list shows some
examples:
●
A = A
●
(A<5) OR (A>=5)
●
0*A = 0
●
2*A = A+A
●
A OR (NOT A)
●
MAX(...) >= ALL (<set of all values>)
●
MIN(...) <= ALL (<set of all values>)
Figure 172: Do these Queries have the Same Result?
114
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Lesson: Understanding NULL Values
Figure 173: The Queries have Different Results!
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115
Unit 6: NULL Values
116
© Copyright. All rights reserved.
Unit 6
Exercise 5
117
Understand NULL Values
Business Example
Your applications using SAP HANA SQL sometimes produce unexpected results. You are
tasks with the analysis of the SELECT statements that have been identified as the potential
cause of these issues.
1. Analyze the following two queries. Do they lead to the same result? Explain your
conclusion.
Query 1:
SELECT * FROM Official
WHERE PNr <> ALL( SELECT PNr FROM Official
WHERE Overtime = Overtime)
Query 2:
SELECT * FROM Official
WHERE Overtime IS NULL
2. Analyze the following two queries. Do they lead to the same result? Explain your
conclusion.
Query 1:
SELECT Name, Overtime FROM Official
WHERE Overtime >= ALL( SELECT Overtime FROM Official)
Query 2:
SELECT Name, Overtime FROM Official
WHERE Overtime = (SELECT MAX(Overtime) FROM Official)
3. Analyze the following two queries. Do they lead to the same result? Explain your
conclusion.
Query 1:
SELECT Name, Overtime FROM Official
WHERE Overtime >= ALL( SELECT Overtime FROM Official
WHERE Overtime IS NOT NULL)
Query 2:
SELECT Name, Overtime FROM Official
WHERE Overtime = (SELECT MAX(Overtime) FROM Official)
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117
Unit 6
Solution 5
118
Understand NULL Values
Business Example
Your applications using SAP HANA SQL sometimes produce unexpected results. You are
tasks with the analysis of the SELECT statements that have been identified as the potential
cause of these issues.
1. Analyze the following two queries. Do they lead to the same result? Explain your
conclusion.
Query 1:
SELECT * FROM Official
WHERE PNr <> ALL( SELECT PNr FROM Official
WHERE Overtime = Overtime)
Query 2:
SELECT * FROM Official
WHERE Overtime IS NULL
a) The two queries have the same result.
b) In the first query, the sub query returns the list of PNrs of all Officials with known
Overtime value. The outer query returns the list of Officials whose PNr is not in the sub
query result, hence the list of Officials with unknown or NULL-valued Overtime. That’s
exactly what the second statement does.
2. Analyze the following two queries. Do they lead to the same result? Explain your
conclusion.
Query 1:
SELECT Name, Overtime FROM Official
WHERE Overtime >= ALL( SELECT Overtime FROM Official)
Query 2:
SELECT Name, Overtime FROM Official
WHERE Overtime = (SELECT MAX(Overtime) FROM Official)
a) The two queries have different results.
b) The first query has an empty result set. No Overtime value is greater or equal to all the
values returned by the sub query since the sub query result also includes NULL values.
The logical expression in the WHERE clause of the outer query thus never evaluates to
true but always to unknown.
c) The aggregate expression in the sub query in the second query ignores NULL values,
hence the outer query determines the Official with the largest Overtime value.
3. Analyze the following two queries. Do they lead to the same result? Explain your
conclusion.
118
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Lesson: Understanding NULL Values
Query 1:
SELECT Name, Overtime FROM Official
WHERE Overtime >= ALL( SELECT Overtime FROM Official
WHERE Overtime IS NOT NULL)
Query 2:
SELECT Name, Overtime FROM Official
WHERE Overtime = (SELECT MAX(Overtime) FROM Official)
a) The two queries have the same result.
b) Since the inner query of the first query now explicitly excludes NULL-valued Overtime
values, the outer query determines the Officials with an Overtime value greater or
equal to all known values. This is equivalent to determine the officials with the
maximum overtime value, hence the second query.
Note:
The case in which all Overtime values are unknown requires special
consideration. In this case, the result of the first query could either be
empty or contain all Officials, depending on how the DBMS handles the ALL
expression with an empty set.
SAP HANA returns an empty result.
© Copyright. All rights reserved.
119
Unit 6: NULL Values
LESSON SUMMARY
You should now be able to:
●
120
Interpret NULL values in databases and understand why their presence can lead to
unexpected query results
© Copyright. All rights reserved.
UNIT 7
Data Stored in Tables
Lesson 1
Changing Data Stored in Tables
Exercise 6: Insert, Update, and Delete Rows in a Database Table
122
129
UNIT OBJECTIVES
●
Add rows to database tables using SQL
●
Change existing rows of a database table
●
Remove existing rows from a database table
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121
Unit 7
Lesson 1
122
Changing Data Stored in Tables
LESSON OVERVIEW
This lesson covers how you can add data to a database, change and delete data.
LESSON OBJECTIVES
After completing this lesson, you will be able to:
●
Add rows to database tables using SQL
●
Change existing rows of a database table
●
Remove existing rows from a database table
Modifying Data — Overview
In the previous lessons, we have examined how you can read data from a database.
Statements to Modify Data — Overview
The following list provides an overview of the statements used to modify data:
INSERT
The INSERT statement adds rows to a database table.
UPDATE
The UPDATE statement changes values of rows existing in a table.
DELETE
The DELETE statement permanently removes existing rows from a table.
UPSERT
The UPSERT changes existing rows of table if found, else inserts new rows. It can be
considered as a combination of INSERT and UPDATE.
REPLACE is a synonym for UPSERT.
Inserting Data
The INSERT statement is used to add new rows to a database table. The DBMS checks If the
table already contains rows with the same primary key values or if you try to add several rows
with the same primary key values, and raises an exception in such a case.
You can use three different syntax versions to insert data. The first two options allow inserting
single rows only, the last option allows inserting several rows with a single statement.
122
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Lesson: Changing Data Stored in Tables
Figure 174: Single Row Insert, Pass Values by Column Order
Figure 175: Single Row Insert, Pass Values by Name
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123
Unit 7: Data Stored in Tables
Figure 176: Omitting Values on Insert
Figure 177: Inserting Multiple Rows
Updating Data
The fundamental statement to update existing rows of a table is the UPDATE statement. Its
basic syntax is
UPDATE <table>
SET <column> = <value> [,
<column> = <value>, ...]
[WHERE <condition>]
124
© Copyright. All rights reserved.
Lesson: Changing Data Stored in Tables
Figure 178: Update Example
Note:
The WHERE clause is optional. But omitting it updates all rows of the table — which
may not be what you intend.
In practice, the WHERE clause often specifies the primary key values of the rows to be updated.
Note that “missing rows” do not lead to an exception, but only to no row being updated.
Figure 179: Updating Several Columns
The WHERE clause supports the same features as for the SELECT statement, for example sub
queries.
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125
Unit 7: Data Stored in Tables
Figure 180: The WHERE Clause Works as Usual
You can also use correlated sub-queries, functions etc. in the WHERE clause.
Figure 181: Sub Queries also Work in the SET Clause
The UPSERT / REPLACE Statement
In addition to the INSERT and the UPDATE statements, SAP HANA also supports an UPSERT
statement. This statement either updates rows in a table or inserts new rows.
UPSERT Statement Syntax
Due to its support for inserting data, the UPSERT statement comes in three syntax variants
similar to the INSERT syntax variants:
UPSERT Table(Column List) VALUES( Value List ) WHERE Condition
If the condition evaluates to true, the matching rows are updated. Else inserts a new row
with the values provided.
UPSERT Table(Column List) VALUES( Value List ) WITH PRIMARY KEY
126
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Lesson: Changing Data Stored in Tables
Updates the row with the primary key contained in the value list if present, else Inserts a
new row with the values provided.
UPSERT Table(Column List) SELECT ... FROM ... WHERE ...
For each row returned by the sub query, checks if the table contains a row with the same
primary key value. If yes updates the existing row with the row returned by the sub query
else inserts a corresponding row.
The first variant looks similar to the INSERT statement, with the key word changed to UPSERT
and a WHERE clause added.
The third variant looks like the INSERT statement in its set variant and behaves mainly the
same.
If you prefer, you can also use the REPLACE keyword alternatively. UPSERT and REPLACE are
truly synonyms.
Deleting Data
The fundamental statement to delete existing rows of a table is the DELETE statement. Its
basic syntax is
DELETE FROM <table>
[WHERE <condition>]
Figure 182: Deleting Selected Rows
Note:
The WHERE clause is optional. But omitting it deletes all rows of the table — which
may not be what you intend.
In practice, the WHERE clause often specifies the primary key values of the rows to be updated.
Note that “missing rows” do not lead to an exception, but only to no row being updated.
The WHERE clause supports the same features as for the SELECT statement, for example sub
queries.
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127
Unit 7: Data Stored in Tables
Figure 183: The WHERE Clause Works as Usual
You can also use correlated sub-queries, functions etc. in the WHERE clause.
128
© Copyright. All rights reserved.
Unit 7
Exercise 6
129
Insert, Update, and Delete Rows in a Database
Table
Business Example
You are asked to use fill a new database table STAFF, with the same set of columns as table
EMPLOYEE, with data.
1. Copy the data from table "EMPLOYEE" into the (structural equal) table "STAFF".
2. Set the remaining vacation days of "Mr A" of table "STAFF" to 50 days.
3. Reduce the remaining vacation days to 18 for all employees (table "STAFF"), who have
more than 18 days.
4. Triple the remaining vacation days of employees (table "STAFF") working for department
"Development" or "Test". Use an uncorrelated sub query with "IN".
5. Create a new entry in table "STAFF".
6. "Mr A" is leaving the company. Delete his data from table "STAFF".
7. Delete employees from table "STAFF", where the department information is unknown.
8. The departments "Planning" and "Test" are closed. Delete all employees of those
departments from table "STAFF". Use an uncorrelated sub query with "IN".
© Copyright. All rights reserved.
129
Unit 7
Solution 6
130
Insert, Update, and Delete Rows in a Database
Table
Business Example
You are asked to use fill a new database table STAFF, with the same set of columns as table
EMPLOYEE, with data.
1. Copy the data from table "EMPLOYEE" into the (structural equal) table "STAFF".
a) INSERT INTO STAFF SELECT * FROM Employee;
2. Set the remaining vacation days of "Mr A" of table "STAFF" to 50 days.
a) UPDATE Staff
SET RemainderDays = 50
WHERE Name = 'Mr A';
3. Reduce the remaining vacation days to 18 for all employees (table "STAFF"), who have
more than 18 days.
a) UPDATE Staff
SET RemainderDays = 18
WHERE RemainderDays > 18;
4. Triple the remaining vacation days of employees (table "STAFF") working for department
"Development" or "Test". Use an uncorrelated sub query with "IN".
a) UPDATE Staff
SET RemainderDays = 3 * RemainderDays WHERE
DepID IN (SELECT DepID
FROM Department WHERE
Name = 'Development' OR Name = 'Test');
5. Create a new entry in table "STAFF".
a) INSERT INTO Staff VALUES ('D200000', 'Mr X', '1977-07-07', 10,
'A01');
6. "Mr A" is leaving the company. Delete his data from table "STAFF".
a) DELETE
FROM Staff
WHERE Name = 'Mr A';
7. Delete employees from table "STAFF", where the department information is unknown.
a) DELETE
FROM Staff
WHERE DepID IS NULL;
8. The departments "Planning" and "Test" are closed. Delete all employees of those
departments from table "STAFF". Use an uncorrelated sub query with "IN".
a) DELETE
FROM Staff WHERE DepID IN (SELECT DepID FROM
Department WHERE Name = 'Planning' OR Name = 'Test');
130
© Copyright. All rights reserved.
Lesson: Changing Data Stored in Tables
LESSON SUMMARY
You should now be able to:
●
Add rows to database tables using SQL
●
Change existing rows of a database table
●
Remove existing rows from a database table
© Copyright. All rights reserved.
131
Unit 7: Data Stored in Tables
132
© Copyright. All rights reserved.
UNIT 8
Data Storage
Lesson 1
Defining How Data is Stored
Exercise 7: Create, Change, and Delete a Database Table
134
143
UNIT OBJECTIVES
●
List the most important data types SAP HANA supports
●
Create new database tables in HANA
●
Change tables by adding, removing or renaming columns
© Copyright. All rights reserved.
133
Unit 8
Lesson 1
134
Defining How Data is Stored
LESSON OVERVIEW
This lesson covers how you can create and change database tables to define how data is
stored.
LESSON OBJECTIVES
After completing this lesson, you will be able to:
●
List the most important data types SAP HANA supports
●
Create new database tables in HANA
●
Change tables by adding, removing or renaming columns
Managing Tables in SAP HANA — Overview
Before data can be inserted into database tables, the tables themselves have to be defined
and created.
Some Options to Manage Database Tables in SAP HANA
SAP HANA supports several means for this purpose, some of which are the following:
●
Using SQL DDL Statements. This is the only option covered in this course.
●
Using a form based editor in the SAP HANA Tools.
●
Using Core Data Services and the HANA Development Perspective in the SAP HANA Tools.
Note:
SQL and the form-based editor provide no support for managing the lifecycle of
database tables, for example for keeping track of how a table was defined or
changed such that changes can be transported to another system without losing
data in the target system.
Nevertheless, this course only covers the SQL option. Understanding it can be
helpful for prototyping or analyzing issues, for example.
SAP HANA SQL Types
When managing database tables on the SQL level, it is important to know the data type
system provided by SAP HANA on the SQL level. This type system includes the following
types.
Table 7: Numerical Data Types
The following table lists the numerical SQL data types supported by SAP HANA
134
© Copyright. All rights reserved.
Lesson: Defining How Data is Stored
Data Type
Remark
Inte- TINYINT
gers
SMALLINT
Real
num
bers
unsigned, 8–bit integer values
signed, 16–bit integer values
INTEGER
signed, 32–bit integer values
BIGINT
signed, 64–bit integer values
DECIMAL(p,s)
Fixed-point decimal number with total number of digits p (precision) and s (scale) digits right of the decimal point (fractional digits).
SMALLDECIMAL
Floating point number with precision varying dynamically between
1 and 16 and scale between –369 and 368, respectively between 1
and 34 and scale between –6111 and 6167
DECIMAL
REAL
32–bit floating point number
DOUBLE
64–bit floating point number
FLOAT(N)
32-bit or 64-bit real number with N significant bits. 1 <= N<= 53.
Note:
DECIMAL(p,s) is suitable for storing fixed-point decimal numbers with p digits in
total and s fractional digits, whereas DECIMAL (without precision and scale) is
used to store floating point numbers.
Since floating point numbers often cannot be represented exactly in the binary
system and are rounded, you should use them with care in database tables.
Especially avoid using them in WHERE clauses.
The number 1.0000259 has precision 8 and scale 7. The number 3.1415 requires at least
precision 5 and scale 4 to be stored without loss of precision. Using DECIMAL(5,4) it is
stored as 3.1415, using DECIMAL(8,7) as 3.1415000.
Table 8: String-/Character Data Types
The following table lists the string—like SQL data types supported by SAP HANA
Data Type
Remark
VARCHAR(n)
Variable-length ASCII character string with maximum length n (n ≤
5,000)
NVARCHAR(n)
Variable-length unicode character string with maximum length n (n ≤
5,000)
ALPHANUM(n)
Variable-length alphanumeric character string with maximum length
n (n ≤ 127)
SHORTTEXT(n)
Variable-length unicode character string based on NVARCHAR(n)
with support for text- and string-search features
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135
Unit 8: Data Storage
Table 9: Data Types for Date and Time
The following table lists the string—like SQL data types supported by SAP HANA
Data Type
Remark
DATE
Consists of year, month, day '2012-05-21'
TIME
Consists of hour, minute, second '18:00:57'
SECONDDATE
Combination of data and time '2012-05-21 18:00:57'
TIMESTAMP
Precision: ten millionth of a second '2012-05-21 18:00:57.1234567'
Table 10: Data Types for Binary Data and Large Objects
The following table lists the string—like SQL data types supported by SAP HANA
Data Type
Remark
Binary
Data
VARBINARY(n)
Binary data, maximum length n Bytes (n ≤ 5,000)
Large
Objects
BLOB
Large binary data (maximum 2 GB)
CLOB
Long ASCII character string (maximum 2 GB)
NCLOB
Long unicode character string (maximum 2 GB)
TEXT
Long unicode character string (maximum 2 GB) based on
NCLOB with support for text- and string-search features
Data type VARBINARY can be used to store short values in binary form, for example UUIDs:
INSERT INTO MyTable VALUES( ... TO_BINARY('Walldorf') ...);
-- Value stored in DB: 57616C6C646F7266
INSERT INTO MyTable VALUES( ... TO_BINARY(x'00075341500700FF') ...);
-- x'..' is a hexadecimal literal value
●
●
136
The term LOB data type (LOB = large object) is used as a generic term for data types such
as CLOB (character large object) or BLOB (binary large object).
When using LOB data types it is important to note the following:
-
LOB columns can not be part of the primary key
-
LOB columns can not be used in the ORDER BY clause
-
LOB columns can not be used in the GROUP BY clause
-
LOB columns may not be part of the JOIN condition (explicit JOIN)
-
LOB columns can not be used as an argument for an aggregate function
-
LOB columns can not be used in the SELECT DISTINCT clause
-
LOB columns can not occur in an UNION statement
-
LOB columns can not be part of a database index
© Copyright. All rights reserved.
Lesson: Defining How Data is Stored
Creating Database Tables
In ANSI SQL, the statement to create new database tables is CREATE TABLE ... SAP HANA
supports two types of table stores — row store and column store. As a result SAP HANA SQL
extends the ANSI syntax to allow controlling whether a new table shall be created in the row
store or in the column store. This leads to the following basic syntax:
Basic CREATE TABLE syntax
CREATE [ROW|COLUMN] TABLE <Table
<Column Name> <SQL Data Type>
<Column Name> <SQL Data Type>
<Column Name> <SQL Data Type>
[PRIMARY KEY (<column list>)]
)
Name> (
[PRIMAY KEY],
[NULL|NOT NULL],
[DEFAULT <Default Value>,
[UNIQUE (<column list>)]
Note:
You do not have to specify which table store to use by selecting the keywords ROW
or COLUMN. The ROW store is the default for reasons of compatibility, but the
COLUMN is the option recommended in most cases.
This is why the following examples all include the keyword COLUMN.
Figure 184: Defining a Table without a Primary Key
Figure 185: Table and Column Names
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137
Unit 8: Data Storage
Figure 186: Defining a Simple Primary Key
Figure 187: Defining a Composite Primary Key
Figure 188: Default Values and Prohibiting NULL Values
138
© Copyright. All rights reserved.
Lesson: Defining How Data is Stored
Figure 189: Simple Uniqueness Constraints
Figure 190: Unique does not Exclude NULL
Figure 191: Multiple Simple Uniqueness Constraints
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139
Unit 8: Data Storage
Figure 192: Composite Uniqueness Constraints
Changing Database Tables
Once a table has been defined, you can modify it in certain ways using the ALTER TABLE
statement. The statement is always followed by the name of the table to be changed, and an
additional keyword to indicate the type of change:
●
ALTER TABLE Table Name
●
-
ADD to add one or more columns
-
DROP
to remove a column (including its data)
-
ALTER
to change properties of columns such as their SQL data type, default value etc.
Figure 193: Adding Columns
140
© Copyright. All rights reserved.
Lesson: Defining How Data is Stored
Figure 194: Removing Columns
Figure 195: Changing Column Properties
To remove a default value, use ALTER TABLE ... ALTER and set the default value to NULL.
Figure 196: Removing the Primary Key
Figure 197: Adding a Primary Key
Renaming a Column
You can also rename an existing table column. This is not achieved using an ALTER TABLE
statement, but using a RENAME COLUMN statement.
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Unit 8: Data Storage
Figure 198: Renaming a Column
Renaming a Table
You can also rename a database table, using a RENAME TABLE statement.
Figure 199: Renaming a Table
Removing a Table
You can also remove an existing table, including its data.
Figure 200: Removing a Table
142
© Copyright. All rights reserved.
Unit 8
Exercise 7
143
Create, Change, and Delete a Database Table
1. Create a new column table "Locations" with the following columns:
Column Name
Data Type
Token
CHAR length 3
Region
NVARCHAR length 20
Country
NVARCHAR length 20
City
NVARCHAR length 20
Each location is uniquely identified by "Token". When inserting data, "Country" and
"Region" can be unknown, but not the "City". If no "Region" is specified during insert the
default value "EMEA" should be used.
2. Create a new column table EmployeeSalary with columns.
Column Name
Data Type
DNumber
NVARCHAR length 7
FromDate
DATE
FixedSalary
Choose a numerical type that can hold salaries between 1000 and 1 million €, and allows for cent precision (two fractional digits).
VariableSalary
Choose a numerical type that can hold percentage values with two fractional digits.
The combination of DNumber and FromDate uniquely identify a row. Neither FixedSalary
nor VariableSalary may be unknown. The VariableSalaray defaults to 10%.
To check your definition, insert three rows into the table, each with a salary valid from the
current date:
●
Employee D100001 with a fixed salary of 2500€ and the default percentage of variable
salary.
●
Employee D100002 with a fixed salary of 3000€ and 15% variable salary,
●
Employee D100003 with a fixed salary of 3200€ and 17% variable salary.
3. Extend the newly created table "Locations" with columns.
• "Population" – type INTEGER
• "Mayor" – type VARCHAR, length 30
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143
Unit 8: Data Storage
4. Remove the columns "Country" and "Population" from table "Locations".
5. Change the table definition of "Locations" that column "City" can have NULL values.
6. Delete tables "Locations" and “EmployeeSalary”.
144
© Copyright. All rights reserved.
Unit 8
Solution 7
145
Create, Change, and Delete a Database Table
1. Create a new column table "Locations" with the following columns:
Column Name
Data Type
Token
CHAR length 3
Region
NVARCHAR length 20
Country
NVARCHAR length 20
City
NVARCHAR length 20
Each location is uniquely identified by "Token". When inserting data, "Country" and
"Region" can be unknown, but not the "City". If no "Region" is specified during insert the
default value "EMEA" should be used.
CREATE
a) COLUMN TABLE Locations (
Token
CHAR(3) PRIMARY KEY,
Region NVARCHAR(20) DEFAULT 'EMEA',
Country NVARCHAR(20),
City
NVARCHAR(20) NOT NULL);
2. Create a new column table EmployeeSalary with columns.
Column Name
Data Type
DNumber
NVARCHAR length 7
FromDate
DATE
FixedSalary
Choose a numerical type that can hold salaries between 1000 and 1 million €, and allows for cent precision (two fractional digits).
VariableSalary
Choose a numerical type that can hold percentage values with two fractional digits.
The combination of DNumber and FromDate uniquely identify a row. Neither FixedSalary
nor VariableSalary may be unknown. The VariableSalaray defaults to 10%.
To check your definition, insert three rows into the table, each with a salary valid from the
current date:
●
Employee D100001 with a fixed salary of 2500€ and the default percentage of variable
salary.
●
Employee D100002 with a fixed salary of 3000€ and 15% variable salary,
●
Employee D100003 with a fixed salary of 3200€ and 17% variable salary.
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145
Unit 8: Data Storage
CREATE
a) COLUMN TABLE EmployeeSalary(
DNumber
NVARCHAR(7),
FromDate
DATE,
FixedSalary
DECIMAL(9,2) NOT NULL,
VariableSalary DECIMAL(5,2) NOT NULL DEFAULT 5.0,
PRIMARY KEY(DNumber,FromDate) );
INSERT INTO
VALUES
INSERT INTO
VALUES
INSERT INTO
VALUES
EmployeeSalary
('D100001', current_date, 2500.0, 5.0 );
EmployeeSalary
('D100002', current_date, 3000.0, 15.0);
EmployeeSalary
('D100003', current_date, 3200.0, 17.0);
3. Extend the newly created table "Locations" with columns.
• "Population" – type INTEGER
• "Mayor" – type VARCHAR, length 30
a) ALTER TABLE LOCATIONS ADD (Population INTEGER, Mayor
VARCHAR(30));
4. Remove the columns "Country" and "Population" from table "Locations".
a) ALTER TABLE LOCATIONS DROP (Country, Population);
5. Change the table definition of "Locations" that column "City" can have NULL values.
a) ALTER TABLE LOCATIONS ALTER (City VARCHAR(20) NULL);
6. Delete tables "Locations" and “EmployeeSalary”.
a) DROP TABLE LOCATIONS;
b) DROP TABLE EMPLOYEESALARY;
146
© Copyright. All rights reserved.
Lesson: Defining How Data is Stored
LESSON SUMMARY
You should now be able to:
●
List the most important data types SAP HANA supports
●
Create new database tables in HANA
●
Change tables by adding, removing or renaming columns
© Copyright. All rights reserved.
147
Unit 8: Data Storage
148
© Copyright. All rights reserved.
UNIT 9
Views For Data Access
Lesson 1
Using Views for Data Access
Exercise 8: Work with Database Views
150
157
UNIT OBJECTIVES
●
Describe the use cases for and advantages of using database views, define database views
and use them in queries
© Copyright. All rights reserved.
149
Unit 9
Lesson 1
150
Using Views for Data Access
LESSON OVERVIEW
This lesson covers database views.
LESSON OBJECTIVES
After completing this lesson, you will be able to:
●
Describe the use cases for and advantages of using database views, define database views
and use them in queries
Database Views — Overview
The following views are supported by SAP HANA:
●
You can define database views using SQL DDL Statements.
This is the only option covered in this course.
●
●
You can define Calculation Views or more generally information models using SAP HANA
Tools.
You can define views using Core Data Services and the HANA Development Perspective in
the SAP HANA Tools.
Note:
Calculation Views and Core Data Servies may be more powerful than SQL views,
or may simplify the development process.
Nevertheless, this course only covers SQL views. Understanding them is the
foundation for understanding the other types of views and can be helpful for
prototyping or analyzing issues.
Database Views Provide the Following Advantages:
●
They allow to de-couple applications from lower levels (relative to the three level schema
architecture)
-
●
You can create views tailored to the users (applications) and their needs
-
●
150
You can use them to change to the conceptual schema (e.g, add columns to a table)
without causing side effects on applications.
The application programmers do not have to know the full database structure but only
the excerpt that is relevant to them.
You can simplify queries
© Copyright. All rights reserved.
Lesson: Using Views for Data Access
-
●
Complex queries referencing views are easier to formulate, provided that the view
provides an appropriate pre-selection of data.
You can use them to limit access to data
-
You can us views to realize value-dependent access privileges or to make sure certain
columns are never visible for .
Figure 201: A Sample View: Red Cars
Figure 202: Using Views in Queries
Managing Database Views
You can create database views using the CREATE VIEW statement. Its syntax basically looks
as follows:
CREATE VIEW <Name of the view> [(<list of column names>)] AS
SELECT ...
You can use nearly any valid SELECT statement as the SELECT clause for a CREATE VIEW
statement.
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151
Unit 9: Views For Data Access
Figure 203: CREATE VIEW
You can also use the following in the SELECT part of the CREATE VIEW statement:
●
●
Calculations and functions
Any type of join. The need to formulate the same complex join condition frequently in
application development is a typical case where views can simplify
●
A TOP N clause to limit the number of rows returned
●
DISTINCT
. Note that a view can contain a projection leading to potential duplicates in result sets
●
Aggregate expressions, GROUP BY and HAVING clauses
●
Sub queries, uncorrelated or correlated
●
UNION
●
152
An ORDER BY clause to achieve a default sort order of results. Note that this order can be
overridden when SELECTing from the view, hence the use of ORDER BY is questionable and
not recommended
© Copyright. All rights reserved.
Lesson: Using Views for Data Access
Figure 204: Renaming Columns with a View, First Option
Figure 205: Renaming Columns with a View, Recommended Option
Figure 206: Precedence Rule for Column Naming
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153
Unit 9: Views For Data Access
Figure 207: Using Aggregation and Grouping is Possible
You can also define views based on other views.
Figure 208: View on View
Figure 209: Removing a View
Modifying Data with a View
In principle, an application developer should not be able to tell if a database table or a view is
accessed when looking at an SQL statement. This largely holds true as far as SELECT
statements / querying data is concerned.
154
© Copyright. All rights reserved.
Lesson: Using Views for Data Access
To a certain extent, this statement even holds true when considering data modifications.
Depending on how a database view is defined, you can insert, update or delete database
records using the view.
Figure 210: Inserting through a View
Figure 211: Updating through a View
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155
Unit 9: Views For Data Access
Figure 212: Deleting through a View
Such data modifications through database views are not always possible. You cannot modify
data through a view your database view definition contains at least one of the following:
156
●
A calculation or function in the projection list
●
A sub query in the projection list
●
DISTINCT
●
The TOP N clause
●
Aggregate expressions
●
A GROUP BY clause
●
UNION
© Copyright. All rights reserved.
Unit 9
Exercise 8
157
Work with Database Views
1. Create view "ManagerView". The manager of department "A04" should only view data of
his own employees. To avoid discrimination/favoritism the birthday should be excluded.
Rename column "RemainderDays" to "VacationDays".
2. Delete view "ManagerView".
© Copyright. All rights reserved.
157
Unit 9
Solution 8
158
Work with Database Views
1. Create view "ManagerView". The manager of department "A04" should only view data of
his own employees. To avoid discrimination/favoritism the birthday should be excluded.
Rename column "RemainderDays" to "VacationDays".
CREATE
a) VIEW ManagerView
AS SELECT DNumber, Name, RemainderDays AS VacationDays
FROM Employee
WHERE DepID = 'A04';
2. Delete view "ManagerView".
a) DROP VIEW ManagerView;
158
© Copyright. All rights reserved.
Lesson: Using Views for Data Access
LESSON SUMMARY
You should now be able to:
●
Describe the use cases for and advantages of using database views, define database views
and use them in queries
© Copyright. All rights reserved.
159
Unit 9: Views For Data Access
160
© Copyright. All rights reserved.
UNIT 10
Functions and Procedures
Lesson 1
Creating User-Defined Functions
Exercise 9: Create and Call Scalar User-Defined Functions
Exercise 10: Create and Call Table User-Defined Functions
162
167
173
Lesson 2
Creating Database Procedures
Exercise 11: Create and Use a Database Procedure
178
185
UNIT OBJECTIVES
●
Create and use scalar and table user-defined functions
●
Create and use database procedures in SAP HANA and know how to debug them
© Copyright. All rights reserved.
161
Unit 10
Lesson 1
162
Creating User-Defined Functions
LESSON OVERVIEW
This lesson introduces SQLScript and explains how to create and use user-defined functions.
LESSON OBJECTIVES
After completing this lesson, you will be able to:
●
Create and use scalar and table user-defined functions
SQLScript Basics
Applications benefit most from the potential of SAP HANA if they perform as many dataintensive computations in the database as possible. This avoids loading large amounts of data
into an application server separate from SAP HANA and to leverage fast column operations,
query optimization and parallel execution. This can be achieved to a certain extent if the
applications use advanced SQL statement, but sometimes you may want to push more logic
into the database than possible using individual SQL statements, or make the logic more
readable and maintainable.
SQLScript has been introduced to assist with this task.
SQL Script — What and Why?
●
●
SQL Script is the following:
-
It is the language to write stored procedures and user-defined functions in SAP HANA
-
It is an extension of ANSI SQL
The main goal is to allow the execution of data intensive calculations inside SAP HANA.
This is helpful for the following reasons:
-
-
It allows to eliminate data transfer between database & application tiers
It allows to execute calculations in the database layer to benefit from fast column
operations, query optimization, and parallel execution
SQLScript Advantages
Compared to plain SQL, SQLScript provides the following advantages:
●
●
162
Using SQLScript, complex logic can be broken down into smaller chunks of code. This
enables modular programming, reuse and a better understandability by functional
abstraction. SQL only allows the definition of SQL views to structure complex queries, and
SQL views have no parameters.
SQLScript supports local variables for intermediate results with implicitly defined types.
With standard SQL, it would be required to define globally visible views even for
intermediate steps.
© Copyright. All rights reserved.
Lesson: Creating User-Defined Functions
●
●
SQL Script has flow control logic such as if—then—else clauses that are not available in
SQL.
Stored procedures can returns multiple results, while a SQL query returns only one result
set.
Scalar Used-Defined Functions
The most simple database object for which you can make use of SQLScript are Scalar UserDefined Functions (Scalar UDFs)
Scalar UDFs allow you to the define functions which take a number of input parameters and
return scalar values. Only expressions are allowed in the body of the UDF, so no table
operations, CE functions or array operations.
Scalar User-Defined Functions Support the Following:
●
●
●
●
●
●
They can have any number of scalar input parameters (primitive SQL types). Input
parameters of table type are not supported.
They can return multiple scalar values.
They can contain expressions within their body. Table and array operations are not
supported.
They can be used in the field list or the WHERE clause of SELECT statements — like built-in
functions.
They are callable via direct assignment in other user-defined functions or stored
procedures (x := my_scalar_func() ).
They must be free of side-effects and do not support any type of SQL statement in their
body.
User-defined functions can be managed in several ways in SAP HANA.
Some Options Used to Define User-Defined Functions in SAP HANA:
●
●
Using SQL DDL Statements. This is the only option covered in this course.
Using the HANA Development Perspective in the SAP HANA Tools and the native HANA
application development model.
Note:
SQL statements alone provide no support for managing the lifecycle of functions,
e.g. for adjusting the function code of an existing function. Nevertheless, this
course only covers the SQL option. This is sufficient to get familiar with the
SQLScript syntax and for prototyping or analyzing issues.
The SQL statement to define user-defined functions is CREATE FUNCTION. The basic syntax
to define a scalar user-defined function looks as follows:
Basic Syntax to Define a Scalar User-Defined Function
CREATE FUNCTION <function name>
(<list of input parameters with
type>)
RETURNS <scalar result parameter name and type>
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163
Unit 10: Functions and Procedures
AS BEGIN
<function body>
END;
Figure 213: A Simple Scalar UDF
User-defined functions — both scalar and table — can also be deleted using an SQL
statement. The sytnax is as simple as follows:
Deleting User-Defined Functions
DROP FUNCTION <function name>;
Note:
The only way to change the body of an existing user-defined function using SQL
statements is to delete the function and re-create it.
You can also use imperative logic in scalar user-defined functions, to the extent this does not
conflict with the statements above. Imperative language constructs allow the developer to
control data and control flow, for example loops, scalar variables and IF-THEN-ELSE
statements.
Imperative Logic in SQLScript
●
●
164
Imperative logic allows you to control the flow of the logic
-
Scalar variable manipulation
-
Branching logic, for example using IF-THEN-ELSE
-
Loops — WHILE and FOR
-
DDL statements and INSERT, UPDATE and DELETE statements
Imperative logic is executed exactly as scripted and procedural. It prevents parallel
processing.
© Copyright. All rights reserved.
Lesson: Creating User-Defined Functions
Note:
DDL and DML are not supported in scalar user-defined functions anyway. They
can be used in table user-defined functions and database procedures.
Figure 214: Imperative Logic in a Scalar UDF
You can find the code shown in the figure in file
HA150/02_SCALAR_UDF_CONVERT_HOURS.sql in the instructor folder to be able to easily
demo it. Vary the second parameter in the SELECT statement to show how an invalid value
leads to unexpected results.
You can also raise exceptions using SQLScript, by using DECLARE ... CONDITION at the
beginning of the SQLScript code to declare a named exception condition and by using SIGNAL
to raise the exception in the code. This looks as follows in a scalar user-defined function.
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165
Unit 10: Functions and Procedures
Figure 215: Raising Exceptions in SQLScript
You can find the code shown in the figure in file
HA150/02_SCALAR_UDF_CONVERT_HOURS.sql in the instructor folder to be able to easily
demo it.
166
© Copyright. All rights reserved.
Unit 10
Exercise 9
167
Create and Call Scalar User-Defined Functions
Business Example
The monetary value of the remaining vacation days of the employees is expected to be
needed in several to-be developed queries and views. To avoid having to repeat the
conversion logic each time, you are asked to create a scalar user-defined function that can be
used in all queries and views for the conversion.
1. Create a scalar user-defined function named Remainderdays_Value. The function
should take an integer-valued number of remaining vacation days as input and convert it
into a monetary value (type DEC(15,2)), assuming each vacation day is worth €300. Test
the function by retrieving the full list of employees with their DNumber and the value of the
remaining working days.
2. You are told that the conversion factor from vacation days into their monetary value in
fact depends on the department an employee works for. Extend your UDF to also take the
Department ID as input and to consider the following conversion factors, using a simple
IF-THEN-ELSE clause:
Department
Conversion Factor
A01
350
A02
310
A03
290
A04
325
NULL (unknown)
NULL
All others
300
© Copyright. All rights reserved.
167
Unit 10
Solution 9
168
Create and Call Scalar User-Defined Functions
Business Example
The monetary value of the remaining vacation days of the employees is expected to be
needed in several to-be developed queries and views. To avoid having to repeat the
conversion logic each time, you are asked to create a scalar user-defined function that can be
used in all queries and views for the conversion.
1. Create a scalar user-defined function named Remainderdays_Value. The function
should take an integer-valued number of remaining vacation days as input and convert it
into a monetary value (type DEC(15,2)), assuming each vacation day is worth €300. Test
the function by retrieving the full list of employees with their DNumber and the value of the
remaining working days.
a) A possible definition of the user-defined function is as follows:
CREATE FUNCTION Remainderdays_Value(im_remainderdays INTEGER)
RETURNS ex_result DEC(15,2)
AS BEGIN
ex_result := :im_remainderdays * 300;
END;
b) The following statement can be used to test the scalar UDF:
SELECT DNumber, Remainderdays_Value(Remainderdays)
FROM Employee;
2. You are told that the conversion factor from vacation days into their monetary value in
fact depends on the department an employee works for. Extend your UDF to also take the
Department ID as input and to consider the following conversion factors, using a simple
IF-THEN-ELSE clause:
Department
Conversion Factor
A01
350
A02
310
A03
290
A04
325
NULL (unknown)
NULL
All others
300
a) Since you cannot alter the existing function, delete it first using the following
statement.
DROP FUNCTION Remainderdays_Value;
168
© Copyright. All rights reserved.
Lesson: Creating User-Defined Functions
b) A possible definition of the user-defined function is as follows:
CREATE FUNCTION Remainderdays_Value(im_remainderdays INTEGER,
im_depid NVARCHAR(3))
RETURNS ex_result DEC(15,2)
AS BEGIN
IF :im_depid = 'A01' THEN -- Planning
ex_result := :im_remainderdays * 350;
ELSEIF :im_depid = 'A02' THEN -- Development
ex_result := :im_remainderdays * 310;
ELSEIF :im_depid = 'A03' THEN -- Test
ex_result := :im_remainderdays * 290;
ELSEIF :im_depid = 'A04' THEN -- Sales
ex_result := :im_remainderdays * 325;
ELSEIF :im_depid IS NULL THEN
ex_result := NULL;
ELSE
ex_result := :im_remainderdays * 300;
END IF;
END;
c) The following statement can be used to test the scalar UDF:
SELECT DNumber, Remainderdays_Value(Remainderdays, DepID)
FROM Employee;
© Copyright. All rights reserved.
169
Unit 10: Functions and Procedures
Table User-Defined Functions
In addition to scalar user-defined functions, SAP HANA supports table user-defined functions.
Table User-defined Functions Support the Following:
●
They can have any number of input parameters
●
They return exactly one table
●
Table operations are allowed within the body
●
They are used in the FROM clause of SELECT statements
●
They must be free of side-effects, i.e. DDL statements or the DML statements INSERT,
UPDATE and DELETE cannot be used in function bodies
The SQL statement to define table user-defined functions also is CREATE FUNCTION. The
difference to the definition of a scalar user-defined function is keyword TABLE for the return
parameter and how the value of the sole return parameter is assigned using keyword RETURN:
Basic Syntax to Define a Table User-Defined Function
CREATE FUNCTION <function name> (<list of input parameters with type>)
RETURNS TABLE [table type|(<list of table column definitions>)]
AS BEGIN
<function body>
RETURN <expression to set return table>
END;
A feature that can be helpful in the context of defining table user-defined functions or stored
procedures is dynamic filtering using built-in table function APPLY_FILTER. This function
allows you to apply a dynamic WHERE clause to a database table or table variable and assign
the result to a table variable.
Figure 216: A Table User-Defined Functions Using Dynamic Filtering
170
© Copyright. All rights reserved.
Lesson: Creating User-Defined Functions
You can find the code shown in the figure in file
HA150/03_TABLE_UDF_CONVERT_OFFICIAL.sql in the instructor folder to be able to easily
demo it.
Exceptions can be raised the same way as in scalar user-defined functions.
Exceptions can also be handled, using DECLARE EXIT HANDLER to define a code block to be
executed in case a certain exception is encountered. RESIGNAL allows to raise a handled
exception again, using an error code in the range for user-defined error codes.
You can also nest BEGIN-END blocks for more advanced exception handling, for example to
make sure values are assigned to variables even in case of an exception, as shown in figure
“Handling exceptions in SQLScript”
Figure 217: Handling Exceptions in SQLScript
You can find the code shown in the figure in file
HA150/03_TABLE_UDF_CONVERT_OFFICIAL.sql in the instructor folder to be able to easily
demo it.
SQLScript supports the notion of array as an indexed collection of elements of a single data
type.
With Arrays in SQLScript, You Can Do the Following:
●
Set and access elements by index using <array>[<index>]
●
Add and remove elements from the ends of an array
●
Concatenate two arrays using CONCAT or ||
●
Convert arrays into a table using UNNEST
●
Convert a table column into an array using ARRAY_AGG
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171
Unit 10: Functions and Procedures
●
Determine the length of the array using CARDINALITY
An example showing how this can be used in a table user-defined function is as follows.
Figure 218: Arrays in SQLScript
You can find the code shown in the figure in file
HA150/03_TABLE_UDF_CONVERT_OFFICIAL.sql in the instructor folder to be able to easily
demo it.
172
© Copyright. All rights reserved.
Unit 10
Exercise 10
173
Create and Call Table User-Defined Functions
Business Example
You are asked to also create a table user-defined function that returns the list of employees,
filtered by caller-specified criteria, together with the monetary value of their remaining
vacation days.
1. Create a table user-defined function named EmployeeRemainderDays_Value. The
function shall take a VARCHAR(1000) parameter as input, pass it to function
APPLY_FILTER and return a filtered list of employees with the monetary value of their
remaining vacation days. Reuse your scalar UDF to calculate the monetary value. The
return table should have the following columns:
Column
Type
DNumber
NVARCHAR(7)
DepID
NVARCHAR(3)
RemainderDays_Value
DEC(15,2)
2. Test your table UDF with the following two values for the input filter parameter: an empty
string, and a string such that only employees of department A02 are returned.
3. Extend your function such that it handles the SQL exceptions 257 and 260, caused by
invalid values passed for the filter parameter.
4. Test your UDF again using the test queries you used before. Also test that an incorrect
filter parameter no longer leads to an exception, e.g. parameter value 'DepID=A02'.
5. Replace the call to your scalar UDF by introducing a local table with the departmentspecific conversion rates from vacation days to monetary value.
6. Delete your table and your scalar UDF again.
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Unit 10
Solution 10
174
Create and Call Table User-Defined Functions
Business Example
You are asked to also create a table user-defined function that returns the list of employees,
filtered by caller-specified criteria, together with the monetary value of their remaining
vacation days.
1. Create a table user-defined function named EmployeeRemainderDays_Value. The
function shall take a VARCHAR(1000) parameter as input, pass it to function
APPLY_FILTER and return a filtered list of employees with the monetary value of their
remaining vacation days. Reuse your scalar UDF to calculate the monetary value. The
return table should have the following columns:
Column
Type
DNumber
NVARCHAR(7)
DepID
NVARCHAR(3)
RemainderDays_Value
DEC(15,2)
a) A possible definition of the user-defined function is as follows:
CREATE FUNCTION EmployeeRemainderDays_Value(im_filter VARCHAR(1000) )
RETURNS TABLE ( DNumber nvarchar(7),
DepID
nvarchar(3),
RemainderDays_Value dec(15,2))
AS BEGIN
lt_employee = APPLY_FILTER("EMPLOYEE",:im_filter);
RETURN
SELECT DNumber, DepID,
RemainderDays_Value(RemainderDays,DepID)
AS RemainderDays_Value
FROM :lt_employee;
END;
2. Test your table UDF with the following two values for the input filter parameter: an empty
string, and a string such that only employees of department A02 are returned.
-- no
a) filter - using empty string as filter value:
SELECT * FROM EmployeeRemainderDays_Value('');
-- filter to employees of department A02 - note the single quotes
SELECT * FROM EmployeeRemainderDays_Value('DepID=''A02''');
-- SELECT * FROM EmployeeRemainderDays_Value('DepID=A02'); -- results
in error
3. Extend your function such that it handles the SQL exceptions 257 and 260, caused by
invalid values passed for the filter parameter.
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Lesson: Creating User-Defined Functions
a) Since the function code cannot be altered, delete the function first.
DROP FUNCTION EmployeeRemainderDays_Value;
b) A possible definition of the user-defined function is as follows:
CREATE FUNCTION EmployeeRemainderDays_Value(im_filter VARCHAR(1000) )
RETURNS TABLE ( DNumber nvarchar(7),
DepID
nvarchar(3),
remainderdays_value dec(15,2))
AS BEGIN
DECLARE lv_filter NVARCHAR(1000) := :im_filter;
-- use a block to make sure the RETURN statement is executed
-- even in case of an exception
BEGIN
DECLARE EXIT HANDLER FOR SQL_ERROR_CODE 257, SQL_ERROR_CODE 260
-- or SQLEXCEPTION
BEGIN
lv_filter := '';
-- APPLY_FILTER needs a scalar variable as second argument,
-- a literal does not work
lt_employee = APPLY_FILTER(EMPLOYEE,:lv_filter);
END;
lt_employee = APPLY_FILTER(EMPLOYEE,:lv_filter);
END;
RETURN
SELECT DNumber, DepID,
RemainderDays_Value(RemainderDays,DepID)
AS remainderdays_value
FROM :lt_employee;
END;
4. Test your UDF again using the test queries you used before. Also test that an incorrect
filter parameter no longer leads to an exception, e.g. parameter value 'DepID=A02'.
-- no
a) filter - using empty string as filter value:
SELECT * FROM EmployeeRemainderDays_Value('');
-- filter to employees of department A02 - note the single quotes
SELECT * FROM EmployeeRemainderDays_Value('DepID=''A02''');
-- incorrect filter value leads to no error (and no filtering)
SELECT * FROM EmployeeRemainderDays_Value('DepID=A02');
5. Replace the call to your scalar UDF by introducing a local table with the departmentspecific conversion rates from vacation days to monetary value.
a) A possible definition of the user-defined function is as follows:
DROP FUNCTION EmployeeRemainderDays_Value;
CREATE FUNCTION EmployeeRemainderDays_Value(im_filter VARCHAR(1000) )
RETURNS TABLE ( DNumber nvarchar(7),
DepID
nvarchar(3),
remainderdays_value dec(15,2))
AS BEGIN
DECLARE lv_filter NVARCHAR(1000) := :im_filter;
-- construct table for "conversion rate"
DECLARE depid NVARCHAR(3) ARRAY;
DECLARE worth DEC(5,2)
ARRAY;
depid[1] := 'A01'; worth[1] := 350;
depid[2] := 'A02'; worth[2] := 310;
depid[3] := 'A03'; worth[3] := 290;
depid[4] := 'A04'; worth[4] := 325;
lt_conversion = UNNEST(:depid,:worth) as (DepID,Worth);
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Unit 10: Functions and Procedures
-- use a block to make sure the RETURN statement is executed
-- even in case of an exception
BEGIN
DECLARE EXIT HANDLER FOR SQL_ERROR_CODE 257, SQL_ERROR_CODE 260
-- or SQLEXCEPTION
BEGIN
lv_filter := '';
-- APPLY_FILTER needs a scalar variable as second argument,
-- a literal does not work
lt_employee = APPLY_FILTER(EMPLOYEE,:lv_filter);
END;
lt_employee = APPLY_FILTER(EMPLOYEE,:lv_filter);
END;
RETURN
SELECT DNumber, e.DepID,
RemainderDays*Worth AS remainderdays_value
FROM :lt_employee AS e LEFT OUTER JOIN :lt_conversion AS c
ON e.depid = c.depid;
END;
6. Delete your table and your scalar UDF again.
DROP
a) FUNCTION EmployeeRemainderDays_Value;
DROP FUNCTION Remainderdays_Value;
176
© Copyright. All rights reserved.
Lesson: Creating User-Defined Functions
LESSON SUMMARY
You should now be able to:
●
Create and use scalar and table user-defined functions
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177
Unit 10
Lesson 2
178
Creating Database Procedures
LESSON OVERVIEW
This lessons expands SQLScript and explains how to create and call stored procedures in SAP
HANA.
LESSON OBJECTIVES
After completing this lesson, you will be able to:
●
Create and use database procedures in SAP HANA and know how to debug them
Database Procedures
The most flexible mechanism to implement data-intensive calculations in SAP HANA and for
which SQLScript can be used are database procedures.
Database procedures can be managed in several ways in SAP HANA.
Some Options to Define Database Procedures in SAP HANA
●
●
Using SQL DDL Statements. This is the only option covered in this course.
Using the HANA Development Perspective in the SAP HANA Tools and the native HANA
application development model.
Note:
SQL statements alone provide no support for managing the lifecycle of
procedures, e.g. for adjusting the code of an existing procedure. Nevertheless, this
course only covers the SQL option. This is sufficient to get familiar with the
SQLScript syntax and for prototyping or analyzing issues,
The SQL statement to define database procedures is CREATE PROCEDURE. The following
example illustrates the basic syntax:
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Lesson: Creating Database Procedures
Figure 219: Input and Output Parameters In Procedures
Caution:
User-defined functions and database procedure share a common namespace.
This means that you cannot have a user-defined function and a database
procedure with the same name.
Database procedures can have several output parameters, and a mix of parameters of scalar
and table types is possible.
They can also have input parameters and parameters which are both input and output. This is
why the keywords IN, OUT and INOUT are necessary in the definition of the signature to
indicate the kind of parameter. IN is the default.
Several optional clauses can be used in the procedure header. The most important ones are
the following:
Elements of a Procedure Header
●
●
●
●
LANGUAGE specifies which programming language the procedure is implemented in.
SQLScript is the default.
SQL SECURITY specifies which user’s privileges apply when the procedure is executed.
Default is DEFINER.
DEFAULT SCHEMA specifies the database schema to be used for unqualified database
object accesses within the procedure body.
READS SQL DATA specifies that the procedure is free of side effects, i.e. does not modify
data in the database.
READS SQL DATA prevents the use of constructs in the procedure body that could lead to
side effects, for example INSERT statements or dynamic SQL using EXEC.
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Unit 10: Functions and Procedures
Using declarative logic only — which implies side-effect free programming — has a very big
advantage:
Declarative Logic in SQLScript
●
The procedure developer defines the what data to select using SELECT or CE functions.
●
The database system determines the how and executes accordingly.
●
Data selection using declarative statements can be parallelized massively.
Figure 220: A Procedure Using Declarative Logic Only
You can find the code shown in the figure in file HA150/04_DBPROC_DEMO.sql in the
instructor folder to be able to easily demo it.
The fact that the procedure uses declarative logic only allows executing the first three queries
and assignments in parallel, because their results are mutually independent.
Recap: Imperative Logic in SQLScript
●
●
Imperative logic allows you to control the flow of the logic
-
Scalar variable manipulation
-
Branching logic, for example using IF-THEN-ELSE
-
Loops — WHILE and FOR
-
DDL statements and INSERT, UPDATE and DELETE statements
Imperative logic is executed exactly as scripted, which is procedural, and prevents parallel
processing.
Database procedures can be deleted using an SQL statement.
180
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Lesson: Creating Database Procedures
Deleting User-Defined Functions
The sytnax is as simple as follows:
DROP PROCEDURE <procedure name>;
Note:
The only way to change the body of an existing procedure using SQL statements is
to delete the procedure and re-create it.
You can find the code shown in the figure in file HA150/04_DBPROC_DEMO.sql in the
instructor folder to be able to easily demo it.
The fact that procedures can have table and scalar output parameters can be used to return
more readable error messages in case of exceptions. The following illustrates an example and
how the original SQL error code and text can be accessed using built-in
variables :SQL_ERROR_CODE and :SQL_ERROR_MESSAGE (note the additional colon).
Figure 221: Using SQL Errors In Exceptions
You can find the code shown in the figure in file HA150/04_DBPROC_DEMO.sql in the
instructor folder to be able to easily demo it.
To call a procedure, use the CALL statement. There are different options depending on where
you call the procedure and for passing parameters.
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Unit 10: Functions and Procedures
Figure 222: Calling Procedures
Arguments for IN parameters of table type can either be physical tables, views or table
variables. The actual value passed for tabular OUT parameters must be ‘?’ when calling a
procedure in the SQL Console.
Calling a procedure WITH OVERVIEW will return one result set that holds the information of
which table contains the result of a particular table output parameter. Scalar outputs are
represented as temporary tables with only one cell. When you pass existing tables to the
output parameters WITH OVERVIEW will insert the result set tuples of the procedure into the
provided tables. When you pass '?' to the output parameters, temporary tables holding the
result sets will be generated. These tables will be dropped automatically once the database
session is closed.
Debugging Database Procedures
Database procedures can be debugged using the SAP HANA Tools. Since the recommended
way for developing database procedures is to use the native SAP HANA application
development model using among others repository packages and the SAP HANA
development perspective in the SAP HANA tools, debugging procedures requires using this
approach to a limited extent.
1. In the SAP HANA tools / SAP HANA Studio, open the SAP HANA development
perspective. You can use menu item Window → Open Perspective ... → Other ... or press
the Open Perspective button in the toolbar. Choose SAP HANA Development in either
case.
2. In the Project Explorer in the navigation frame on the left, open the context menu and
choose New ,,, → Other .... In the dialog window, open folder SAP HANA → Application
Development, choose item XS Project and press Next.
3. Enter a Project Name, e.g. debug and press Next.
4. In step Share Project, press Add Workspace, choose your SAP HANA System (connection)
and press Finish.
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Lesson: Creating Database Procedures
5. Back in step Share Project, unflag Add Project Folder as Subpackage.
6. Press Browse ..., choose a repository package, press OK and then Next.
7. In step Create Objects, unflag all checkboxes in the Acces Objects frame and leave the
fields in frame Common Objects empty.
This is important to avoid creating unwanted repository content.
8. Press Finish.
You now have a project in the Project Explorer.
9. Open the project, sub-folder SAP HANA System Library → Catalog to locate the procedure
you want to debug in your database schema.
10. Double-click the procedure to open it in an editor window.
11. In the left-most column of the editor window, double-click on the lines where you want to
set break points.
12. Open the Debug perspective. You can use menu item Window → Open
Perspective ... → Other ... or press the Open Perspective button in the toolbar. Choose
Debug in either case.
13. Open the menu of the Debug button in the toolbar and choose Debug Configurations ...
14. Choose SAP HANA Stored Procedures, open the context menu and choose New.
15. Enter a name for the Debug Configuration.
16. Choose option Procedure to Debug: and Catalog Schemas in the corresponding dropdown
list.
17. Press the Browse ... button, navigate to your database schema and the procedure you
want to debug, choose it and press OK.
18. Press Debug.
The system starts a debugging session and calls the procedure chose in the debug
configuration. Execution stops at the first breakpoint.
You can resume execution until the next breakpoint, terminate the debugging session, inspect
the variable values, add break points, and more.
Note:
Make sure you always finish the execution of the procedure or explicitly terminate
the debugging session when done. Dangling debugging sessions consume
considerable resources, also on the server side.
Debugging Database Procedures
Demonstrate the procedure for debugging database procedures, using the
Convert_OfficialsHours procedure.
1. In case you haven’t done so yet, create the procedure Convert_OfficialsHours as
shown on figures “Input and Output Parameters In Procedures” and “Using SQL Errors In
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183
Unit 10: Functions and Procedures
Exceptions”. You can find the code in file HA150/04_DBPROC_DEMO.sql in the instructor
folder, or use the following code.
CREATE PROCEDURE Convert_OfficialsHours(
IN im_filter VARCHAR(1000),
IN im_to
VARCHAR(1),
OUT ex_official TABLE (PNr
nvarchar(3),
Name
nvarchar(20),
Overtime dec(5,2)),
OUT ex_message NVARCHAR(200))
LANGUAGE SQLSCRIPT
SQL SECURITY INVOKER
READS SQL DATA
AS BEGIN
DECLARE lv_filter NVARCHAR(1000) := :im_filter;
-- use a block to make sure the RETURN statement is executed even in
case of an exception
BEGIN
DECLARE EXIT HANDLER FOR SQL_ERROR_CODE 257, SQL_ERROR_CODE 260 -or SQLEXCEPTION
BEGIN
lv_filter := '';
-- APPLY_FILTER needs a scalar variable as second argument, a
literal does not work
lt_official = APPLY_FILTER(Official,:lv_filter);
ex_message := 'Result not filtered because of SQL ERROR ' ||
::SQL_ERROR_CODE || ' ' || ::SQL_ERROR_MESSAGE;
END;
lt_official = APPLY_FILTER(Official,:lv_filter);
END;
ex_official = SELECT PNr, Name, Convert_Hours(Overtime,:im_to) as
Overtime
FROM :lt_official;
END;
2. To avoid having to enter input parameter values when the procedure is called, first create
a very simple wrapper. The wrapper also illustrates calling a procedure within another
procedure.
CREATE PROCEDURE tester()
AS BEGIN
DECLARE lv_message NVARCHAR(200);
CALL Convert_OfficialsHours( im_filter=>NULL, im_to=>'d',
ex_official=>lt_official,
ex_message=>lv_message );
END;
3. Follow the procedure as detailed above. You can use repository package
workshop.exercises in the Share Project step. Make sure you unflag Add Project
Folder as Subpackage in the same step.
184
© Copyright. All rights reserved.
Unit 10
Exercise 11
185
Create and Use a Database Procedure
Business Example
You are asked to implement a conversion from the employees’ remaining vacation time to
monetary value, this time as a database procedure to allow for flexible access to the logic.
1. Create a database procedure named EmployeeRemainderDays_Value. The procedure
shall take a VARCHAR(1000) parameter as input and use function APPLY_FILTER to only
work on a subset of employee records. The procedure shall return the filtered list of
employees with the monetary value of their remaining vacation days. The return table
should have the following columns:
Column
Type
DNumber
NVARCHAR(7)
DepID
NVARCHAR(3)
RemainderDays_Value
DEC(15,2)
2. Call your procedure with the following two values for the input filter parameter: an empty
string, and a string such that only employees of department A02 are returned. Also test
that an exception caused by an incorrect filter parameter value is caught.
3. Extend the procedure such that it returns a parameter of type NVARCHAR(200) in
addition. The parameter shall contain an error message in case the procedure encounters
an SQL error 257 or 260. The error message shall contain the SQL error code and error
message.
4. Test your procedure again using the test queries you used before.
© Copyright. All rights reserved.
185
Unit 10
Solution 11
186
Create and Use a Database Procedure
Business Example
You are asked to implement a conversion from the employees’ remaining vacation time to
monetary value, this time as a database procedure to allow for flexible access to the logic.
1. Create a database procedure named EmployeeRemainderDays_Value. The procedure
shall take a VARCHAR(1000) parameter as input and use function APPLY_FILTER to only
work on a subset of employee records. The procedure shall return the filtered list of
employees with the monetary value of their remaining vacation days. The return table
should have the following columns:
Column
Type
DNumber
NVARCHAR(7)
DepID
NVARCHAR(3)
RemainderDays_Value
DEC(15,2)
a) Note that the procedure looks quite similar to the table user-defined function from an
earlier exercise. Since the procedure shall have the same name as the earlier function,
first delete the function in case you have not done so yet.
DROP FUNCTION EmployeeRemainderDays_Value;
b) You can largely re-use the CREATE statement of the table function, replacing the
keyword FUNCTION with PROCEDURE, adjusting the definition of the signature and how
the output parameter value is determined. A possible definition of the procedure is as
follows:
CREATE PROCEDURE EmployeeRemainderDays_Value(
in im_filter VARCHAR(1000),
out ex_employee TABLE ( DNumber nvarchar(7),
DepID
nvarchar(3),
remainderdays_value dec(15,2))
)
AS BEGIN
DECLARE lv_filter NVARCHAR(1000) := :im_filter;
-- construct table for "conversion rate"
DECLARE depid NVARCHAR(3) ARRAY;
DECLARE worth DEC(5,2)
ARRAY;
depid[1] := 'A01'; worth[1] := 350;
depid[2] := 'A02'; worth[2] := 310;
depid[3] := 'A03'; worth[3] := 290;
depid[4] := 'A04'; worth[4] := 325;
lt_conversion = UNNEST(:depid,:worth) as (DepID,Worth);
-- use a block to make sure the RETURN statement is executed
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Lesson: Creating Database Procedures
-- even in case of an exception
BEGIN
DECLARE EXIT HANDLER FOR SQL_ERROR_CODE 257, SQL_ERROR_CODE 260
BEGIN
lv_filter := '';
-- APPLY_FILTER needs a scalar variable as second argument,
-- a literal does not work
lt_employee = APPLY_FILTER(EMPLOYEE,:lv_filter);
END;
lt_employee = APPLY_FILTER(EMPLOYEE,:lv_filter);
END;
ex_employee =
SELECT DNumber, e.DepID,
RemainderDays*Worth AS remainderdays_value
FROM :lt_employee AS e LEFT OUTER
JOIN :lt_conversion AS c
ON e.depid = c.depid;
END;
2. Call your procedure with the following two values for the input filter parameter: an empty
string, and a string such that only employees of department A02 are returned. Also test
that an exception caused by an incorrect filter parameter value is caught.
-- no
a) filter - using empty string as filter value:
CALL EmployeeRemainderDays_Value('',?);
-- filter to employees of department A02 - note the single quotes
CALL EmployeeRemainderDays_Value('DepID=''A02''',?);
-- incorrect filter value leads to no error (and no filtering)
CALL EmployeeRemainderDays_Value('Nonsense',?);
3. Extend the procedure such that it returns a parameter of type NVARCHAR(200) in
addition. The parameter shall contain an error message in case the procedure encounters
an SQL error 257 or 260. The error message shall contain the SQL error code and error
message.
a) Note that you cannot simply extend the body of the procedure using SQL. You have to
delete the procedure and re-create it. To delete it use the following statement.
DROP PROCEDURE EmployeeRemainderDays_Value;
b) A possible definition of the procedure is as follows:
CREATE PROCEDURE EmployeeRemainderDays_Value(
IN im_filter VARCHAR(1000),
OUT ex_employee TABLE ( DNumber nvarchar(7),
DepID
nvarchar(3),
Remainderdays_Value dec(16,2)),
OUT ex_message NVARCHAR(200))
AS BEGIN
DECLARE lv_filter NVARCHAR(1000) := :im_filter;
-- construct table for "conversion rate"
DECLARE depid NVARCHAR(3) ARRAY;
DECLARE worth DEC(5,2)
ARRAY;
depid[1] := 'A01'; worth[1] := 350;
depid[2] := 'A02'; worth[2] := 310;
depid[3] := 'A03'; worth[3] := 290;
depid[4] := 'A04'; worth[4] := 325;
lt_conversion = UNNEST(:depid,:worth) as (DepID,Worth);
-- use a block to make sure the RETURN statement is executed
-- even in case of an exception
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Unit 10: Functions and Procedures
BEGIN
DECLARE EXIT HANDLER FOR SQL_ERROR_CODE 257, SQL_ERROR_CODE 260
BEGIN
lv_filter := '';
-- APPLY_FILTER needs a scalar variable as second argument,
-- a literal does not work
lt_employee = APPLY_FILTER(EMPLOYEE,:lv_filter);
ex_message := 'Result not filtered because of SQL ERROR ' ||
::SQL_ERROR_CODE || ' ' || ::SQL_ERROR_MESSAGE;
END;
lt_employee = APPLY_FILTER(EMPLOYEE,:lv_filter);
END;
ex_employee = SELECT DNumber, e.DepID,
RemainderDays*Worth AS remainderdays_value
FROM :lt_employee AS e LEFT OUTER
JOIN :lt_conversion AS c
ON e.depid = c.depid;
END;
4. Test your procedure again using the test queries you used before.
-- no
a) filter - using empty string as filter value:
CALL EmployeeRemainderDays_Value('',?,?);
-- filter to employees of department A02 - note the single quotes
CALL EmployeeRemainderDays_Value('DepID=''A02''',?,?);
-- incorrect filter value leads to no error (and no filtering)
-- Should now return an error message in addition
CALL EmployeeRemainderDays_Value('Nonsense',?,?);
188
© Copyright. All rights reserved.
Lesson: Creating Database Procedures
LESSON SUMMARY
You should now be able to:
●
Create and use database procedures in SAP HANA and know how to debug them
© Copyright. All rights reserved.
189
Unit 10: Functions and Procedures
190
© Copyright. All rights reserved.
UNIT 11
Data Access
Lesson 1
Defining Data Access
192
UNIT OBJECTIVES
●
●
Understand database schemas and access tables in other schemas
Explain when database indexes make sense in SAP HANA and create and delete indexes
using SQL
© Copyright. All rights reserved.
191
Unit 11
Lesson 1
192
Defining Data Access
LESSON OVERVIEW
This lesson covers how you can manage data access permission on SQL level, and how data
access can be sped up using database indexes.
LESSON OBJECTIVES
After completing this lesson, you will be able to:
●
●
Understand database schemas and access tables in other schemas
Explain when database indexes make sense in SAP HANA and create and delete indexes
using SQL
Database Schemas
You created database tables or views in earlier exercises. You may wonder why this did not
lead to naming collisions with other database users creating tables or views with the same
names. You can consider the following situation:
●
User 1 creates table “Car”.
●
User 2 also creates table “Car”.
Why are there no naming collision?
The reason is that the name of the database object (table, view etc.) implicitly contains a
schema name as prefix: <schema name>.<given Name>. Thus the two Car tables are not the
same: USER1.Car ≠ USER2.Car
Database Schemas
●
192
A database schema is the container that holds database objects such as tables or
database views.
●
Each database objects belongs to a schema.
●
A database schema acts as namespace for the database objects it contains.
© Copyright. All rights reserved.
Lesson: Defining Data Access
Figure 223: User Schemas
Figure 224: Queries Across Schemas
Controlling Data Access
Note:
Because the SAP HANA platform is more than a relational database management
system, it also provides a richer authorization model extending beyond the SQL
level. This model includes, for example, analytic privileges, application,
development and administration permissions and roles. SAP HANA therefore also
provides more advanced means to manage user permissions and roles than
covered in the following, for example using the SAP HANA Tools user and role
management.
This course only covers the SQL level and how permissions can be managed using
SQL statements. This can be helpful for prototyping, for example.
Relational database management system, including SAP HANA, support two basic ways for
controlling which database user can access which data using SQL:
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193
Unit 11: Data Access
●
You can create database views that represent a portion of the data.
-
●
Example: You provide a view that contains only the owners of CityA to a user for whom
the owners outside of CityA are not relevant.
Grant specific access permissions to selected users using the Data Control Language of
SQL
-
Example: You grant read only permissions to a user, who should be able to read the
owner data but not to change it.
The Following are the Two DCL Statements and their Basic Syntax:
GRANT
The GRANT statement is used to selectively grant access permissions to a user or role. Its
basic syntax is:
GRANT <list of privileges> ON <database object> TO <user or role> [WITH GRANT
OPTION]
REVOKE
The REVOKE statement is used to selectively withdraw access permissions from a user or
role. Its basic syntax is:
REVOKE <list of privileges> ON <database object> FROM <user or role>
Figure 225: Example: Granting Read Permission
Figure 226: Granting Several Privileges in One Statement
194
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Lesson: Defining Data Access
Note:
There is no inherent dependency between privileges. For example, an INSERT
privilege does not imply a SELECT or an UPDATE privilege.
Figure 227: Granting Access to all Objects of a Schema
Figure 228: Revoking a Privilege
Figure 229: Revoking Multiple Privileges
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195
Unit 11: Data Access
Figure 230: No Hierarchy of Privileges
Database Indexes
A Database Index has the Following Properties:
●
It can be considered an access path that helps the DBMS locate rows faster.
●
It can therefore speed up searching and sorting.
-
It reduces the need for “full table scans” or “full column scans”.
-
In SAP HANA, non-unique indexes are needed in exceptional cases only.
●
It has no influence on the result of queries — only on how fast the result is returned.
●
It can slow down INSERTs, UPDATEs and DELETEs (especially non-unique indexes).
Figure 231: HANA: Why Indexes on Column Store?
196
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Lesson: Defining Data Access
Figure 232: HANA: Table Scan With Value ID
Figure 233: HANA: Inverted Index on Column Store
Managing Database Indexes
The following are the two statements to manage indexes using SQL and their basic syntax:
CREATE INDEX
This statement is used to create an index. Its basic syntax is:
CREATE [UNIQUE] INDEX <index name> ON <table name> ( <list of columns> )
DROP INDEX
This statement is used to delete an index. The underlying database table and its data are
not deleted. The basic syntax is:
DROP INDEX <index name>
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Unit 11: Data Access
Figure 234: Example: Creating a Non-Unique Index
Figure 235: Example: Creating a Unique Index
Creating a unique index is equivalent to adding a UNIQUE constraint to the definition of the
database table. Each such constraint automatically leads to a unique index. In addition, a
unique index is created automatically for the primary key.
You can also do the Following:
●
You can create multiple indexes on the same table.
●
You can create multi-column indexes.
Note:
This is not recommended for non-unique indexes on column store tables in
SAP HANA because of the additional memory overhead involved. Typically,
creating a non-unique index on one, most important column is sufficient to
achieve good performance with SAP HANA.
Figure 236: DROP INDEX
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Lesson: Defining Data Access
LESSON SUMMARY
You should now be able to:
●
●
Understand database schemas and access tables in other schemas
Explain when database indexes make sense in SAP HANA and create and delete indexes
using SQL
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Unit 11: Data Access
200
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UNIT 12
Database Transactions
Lesson 1
Explaining Database Transactions
202
UNIT OBJECTIVES
●
Explain database transactions and the ACID requirements
●
Finish database transactions in HANA using SQL statements
●
Describe issues that arise if transactions are not mutually isolated
●
Understand and control Isolation levels of transactions and how SAP HANA handles
concurrency
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Unit 12
Lesson 1
202
Explaining Database Transactions
LESSON OVERVIEW
This lesson covers the necessity for database transactions and how SAP HANA treats them.
LESSON OBJECTIVES
After completing this lesson, you will be able to:
●
Explain database transactions and the ACID requirements
●
Finish database transactions in HANA using SQL statements
●
Describe issues that arise if transactions are not mutually isolated
●
Understand and control Isolation levels of transactions and how SAP HANA handles
concurrency
Database Transactions
One of the strengths of relational DBMs is their support for database transactions.
Transactions serve the following purpose:
Figure 237: The Purpose of Database Transactions
202
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Lesson: Explaining Database Transactions
Figure 238: The ACID Requirements
Starting and Ending Transactions
Starting a Transaction in SAP HANA
●
●
SAP HANA does not provide a SQL statement to explicitly start a transaction.
A transaction is started implicitly with the first DML statement, before execution of the
statement begins.
A Transaction in SAP HANA can be Ended with the Following Statements:
COMMIT [WORK]
Ends a transaction successfully, making modifications performed during the transaction
durable.
ROLLBACK [WORK]
Ends a transaction by discarding all modifications performed during the transaction, this
way cancels a transaction.
Figure 239: Example Database Transactions
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Unit 12: Database Transactions
DDL Statements automatically end the transaction they belong to successfully, that using an
implicit COMMIT.
Figure 240: Database Transactions
Isolation Challenges
The Isolation requirement becomes relevant when considering concurrent access to a
database by several users or at least sessions.
Support for such multi-user mode is absolutely necessary. Executing all transactions strictly
in sequence would be a huge waste of resources, e.g. when waiting for user input or the
applications using the DBMS. A strict serialization of all transactions also would not be
accepted by the end users.
Several transactions must thus run time-shared. This makes the isolation requirement nontrivial, because without certain control mechanisms, the following five types of problems can
occur.
Figure 241: Isolation Challenge 1: Inconsistent Reads
204
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Lesson: Explaining Database Transactions
Figure 242: Isolation Challenge 2: Lost Updates
Figure 243: Isolation Challenge 3: Phantom Reads
Figure 244: Isolation Challenge 4: Non-Repeatable Reads
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Unit 12: Database Transactions
Figure 245: Isolation Challenge 5: Dirty Reads
Isolation Levels and MVCC
DBMSs, including SAP HANA, allow you to control the extent to which problems are avoided
for each transaction by supporting different Isolation Levels. Isolation levels define whether
rows that are read, inserted, updated, or deleted by a transaction can be accessed by
concurrent transactions and whether the current transaction should be able to access rows
that were read, inserted, updated or deleted by other transactions.
In summary, isolation levels control which types of problems that come from the multi-user
mode are allowed and which ones are prevented.
Figure 246: Isolation Levels
Typical DBMSs support a single version of each persistent data record only and use both read
and write locks (set automatically by the DBMS) to implement the isolation levels. The
isolation level then impacts the granularity of read and write locks.
In contrast, SAP HANA implements Multi-Version Concurrency Control (MVCC) and works
with snapshot isolation. This means that consistency of write access is ensured using
exclusive write locks “as usual, but consistency of read access works as follows:”
206
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Lesson: Explaining Database Transactions
Consistency of Read Access in SAP HANA
●
No shared (read) locks are necessary.
●
A SELECT statement sees a consistent version of the database state — a snapshot.
-
By default, this is the version valid when the transaction the statement belongs to
began — transaction level snapshot isolation.
This roughly corresponds to Isolation Level REPEATABLE READ.
-
This can be changed to the version valid immediately before statement execution —
statement level snapshot isolation.
This roughly corresponds to Isolation Level READ COMMITTED.
●
Write access creates a new version of the database state, which is not visible to other
transactions that started already.
Controlling the Isolation Level
●
The higher the Isolation Level, the more types of problems are prevented.
●
But the higher the burden on the system and the less concurrency.
●
→ Isolation level can be controlled
●
SET TRANSACTION ISOLATION LEVEL IsolationLevel
Figure 247: Isolation Levels Supported by SAP HANA
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Unit 12: Database Transactions
Figure 248: Multi-Version Concurreny Control - An Example
The figure, Multi-Version Concurreny Control - An Example, illustrates that “transaction level
isolation” is not exactly the same as isolation-level “repeatable read” in single-version
concurrency control: In classical DBMSs, T2 and T4 would see V2 if they read it first after T3 is
committed.
Figure 249: Write Locks & Snapshot Isolation - Additional Remarks
LESSON SUMMARY
You should now be able to:
●
Explain database transactions and the ACID requirements
●
Finish database transactions in HANA using SQL statements
●
Describe issues that arise if transactions are not mutually isolated
●
208
Understand and control Isolation levels of transactions and how SAP HANA handles
concurrency
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