Chapter 5 6e & 8 5e: Complex SQL
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Prof. Steven A. Demurjian, Sr.
Computer Science & Engineering Department
The University of Connecticut
191 Auditorium Road, Box U-155
Storrs, CT 06269-3155
steve@engr.uconn.edu
http://www.engr.uconn.edu/~steve
(860) 486 - 4818
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About one third of these slides are being used with the permission of Dr. Ling
Lui, Associate Professor, College of Computing, Georgia Tech.
About one-half of these slides have been adapted from the AWL web site for
the textbook.
Chapter 5-1
Variety of Complex SQL Queries
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Nested Queries
Grouping and Aggregation
Order by and Having
Views
Sets
Tuple Variable
Other Complex Queries and Operations
Chapter 5-2
Recall Earlier Query 1
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Query 1: Retrieve Name and Address of all Employees
who work for the 'Research' Department
SELECT FNAME, MINIT, LNAME, ADDRESS, DNAME
FROM
EMPLOYEE, DEPARTMENT
WHERE DNAME='Research' AND DNUMBER=DNO
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What Action is Being Performed?
Chapter 5-3
Nested Queries
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SQL SELECT Nested Query is Specified within
WHERE-clause of another Query (the Outer Query)
Query 1A: Retrieve the Name and Address of all
Employees who Work for the 'Research' Department
SELECT FNAME, LNAME, ADDRESS
FROM EMPLOYEE
WHERE DNO IN
(SELECT DNUMBER
FROM
DEPARTMENT
WHERE DNAME='Research' )
Note: This Reformulates Earlier Query 1 (not in book)
Chapter 5-4
How Does Nested Query Work?
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The Nested Query Selects Number of 'Research' Dept.
The Outer Query Selects an EMPLOYEE Tuple If Its
DNO Value Is in the Result of Either Nested Query
IN represents Set Inclusion of Result Set
We Can Have Several Levels of Nested Queries
SELECT FNAME, LNAME, ADDRESS
FROM EMPLOYEE
Inner Query Returns
WHERE DNO IN
Relation of DNUMBER
(SELECT DNUMBER
D1, D2, etc.
FROM
DEPARTMENT
WHERE Dname=’Research' )
Chapter 5-5
Correlated Nested Queries
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When WHERE-clause of a Nested Query References
an Attribute of a Relation Declared in the Outer Query
Query 16: Retrieve the Name of each Employee who
has a Dependent with the Same First Name as the
Employee
SELECT E.FNAME, E.LNAME
Inner Query Returns
FROM
EMPLOYEE AS E
Relation a set of
WHERE E.SSN IN
ESSNs for All Emps
(SELECT
ESSN
that have Dependents
FROM
DEPENDENT with the same FNAME
WHERE
ESSN=E.SSN AND
E.FNAME=DEPENDENT_NAME)
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Note: This Differs Slightly from 16 in book.
Chapter 5-6
Query Equivalence
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Query 16:
SELECT E.FNAME, E.LNAME
FROM
EMPLOYEE AS E
WHERE E.SSN IN
(SELECT
ESSN
FROM
DEPENDENT
WHERE
ESSN=E.SSN AND
E.FNAME=DEPENDENT_NAME)
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Query 16A:
SELECT E.FNAME, E.LNAME
FROM
EMPLOYEE E, DEPENDENT D
WHERE E.SSN=D.SSN AND
E.FNAME=D.DEPENDENT_NAME
Chapter 5-7
EXISTS Nested Queries
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EXISTS checks Whether the Result of a Correlated
Nested Query is Empty (contains no tuples) or not
Query 16B: Retrieve the Name of each Employee who
has a Dependent with the Same First Name as the
Employee
Inner Query Returns
SELECT FNAME, LNAME
Is True if there is
FROM
EMPLOYEE
At least one match.
WHERE EXISTS
(SELECT
*
Can there be 2 matches?
FROM
DEPENDENT
WHERE
SSN=ESSN AND
FNAME=DEPENDENT_NAME)
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There is a analogous NOT EXISTS
Chapter 5-8
NULLS in SQL Queries
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SQL Allows Queries that Check if a value is NULL
(Missing or Undefined or not Applicable)
SQL uses IS or IS NOT to compare NULLs since it
Considers each NULL value Distinct from other NULL
Values, so Equality Comparison is not Appropriate
Query 18: Retrieve the names of all employees who do
not have supervisors.
SELECT FNAME, LNAME
FROM
EMPLOYEE
WHERE SUPERSSN IS NULL
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Why Would Such a Capability be Useful?
 Downloading/Crossloading a Database
 Promoting a Attribute to PK/FK
Chapter 5-9
Aggregate Functions in SQL Queries
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Query 19: Find Maximum Salary, Minimum Salary,
and Average Salary among all Employees
SELECT MAX(SALARY), MIN(SALARY),
AVG(SALARY)
FROM
EMPLOYEE
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Query 20: Find maximum and Minimum Salaries
among 'Research' Department Employees
SELECT MAX(SALARY), MIN(SALARY)
FROM
EMPLOYEE, DEPARTMENT
WHERE DNAME='Research' AND DNUMBER=DNO
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What Does Query 22 Do?
SELECT COUNT(*)
FROM
EMPLOYEE, DEPARTMENT
WHERE DNAME='Research' AND DNUMBER=DNO
Chapter 5-10
Grouping in SQL Queries
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In Many Cases, We Want to Apply the Aggregate
Functions to Subgroups of Tuples in a Relation
Each Subgroup of Tuples is Set of Tuples that Have the
Same Value for the Grouping Attribute(s)
Function is Applied to Each Subgroup Independently
Query 24: For Each Department, Retrieve the DNO,
Number of Employees, and Their Average Salary
SELECT DNO, COUNT (*), AVG (SALARY)
FROM EMPLOYEE
GROUP BY
DNO
Chapter 5-11
Grouping in SQL Queries
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Query 24: For Each Department, Retrieve the DNO,
Number of Employees, and Their Average Salary
SELECT DNO, COUNT (*), AVG (SALARY)
FROM EMPLOYEE
GROUP BY
DNO
EMPLOYEE tuples are Divided into Groups; each
group has the Same Value for Grouping Attribute DNO
COUNT and AVG functions are applied to each Group
of Tuples Aeparately
SELECT-clause Includes only the Grouping Attribute
and the Functions to be Applied on each Tuple Group
Chapter 5-12
Results of Query 24:
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SELECT DNO, COUNT (*), AVG (SALARY)
FROM EMPLOYEE
GROUP BY
DNO
Chapter 5-13
Joins and Grouping in SQL Queries
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A Join Condition can be used in Conjunction with
Grouping
Query 25: For each Project, Retrieve its Number,
Name, and Number of Employees working on Project
SELECT
PNUMBER, PNAME, COUNT (*)
FROM
PROJECT, WORKS_ON
WHERE
PNUMBER=PNO
GROUP BY
PNUMBER, PNAME
In this case, the Grouping and Functions are Applied
after the Joining of the two Relations
Chapter 5-14
The HAVING Clause in SQL Queries
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In Some Cases, we want to retrieve values of Functions
for only those Groups that Satisfy Certain Condition(s)
The HAVING-clause is used for Specifying a Selection
Condition on Groups (rather than Individual Tuples)
Query 26: For each Project on which more than two
employees work, Retrieve its Number, Name, and
Number of Employees working on Project project
SELECT
PNUMBER, PNAME, COUNT (*)
FROM
PROJECT, WORKS_ON
WHERE
PNUMBER=PNO
GROUP BY
PNUMBER, PNAME
HAVING
COUNT (*) > 2
Chapter 5-15
Results of Query 26:
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After Applying the WHERE/GROUP BY Clauses
Two Groups Not Selected Based
on Having Constraint
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Chapter 5-16
Results of Query 26:
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After Applying the HAVING Clause Condition
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3
3
3
3
Chapter 5-17
Substring Comparison in SQL Queries
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In Regard to Strings, Most DBMSs Support SQL
Queries for Exact, Near, and Starts with Matching
LIKE is Used to Compare Partial Strings
'%' (or '*') Replaces an Arbitrary # of characters
'_' replaces a single arbitrary character
Query 12: Retrieve all Employees whose Address is in
Houston, Texas.
SELECT FNAME, LNAME
FROM
EMPLOYEE
WHERE ADDRESS LIKE '%Houston,TX% '
Houston, TX can be anywhere within the ADDRESS
VAR CHAR String
Chapter 5-18
Substring Comparison in SQL Queries
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The LIKE Operator Allows us to get Around the Fact
that each Value is Considered Atomic and Indivisible
SQL: Character String Attribute values are not Atomic
Query 12A: Retrieve all employees who were born
during the 1950s.
SELECT
FNAME, LNAME
FROM
EMPLOYEE
WHERE
BDATE LIKE
' __5_______'
There are two “_” before 5 and seven “_” after 5
Chapter 5-19
Arithmetic Operations in SQL Queries
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Standard Arithmetic Operators '+', '-'. '*', and '/' can be
Applied to Numeric Values in an SQL Query Result
Query 13: Show the Effect of Giving all Employees
who work on the 'ProductX' project a 10% raise.
SELECT FNAME, LNAME, 1.1*SALARY
FROM
EMPLOYEE, WORKS_ON, PROJECT
WHERE SSN=ESSN AND PNO=PNUMBER
AND PNAME='ProductX'
Chapter 5-20
ORDER BY Clause in SQL Queries
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ORDER BY used to Sort the Tuples in a Query Result
based on the Values of one or More Attribute(s)
Query 15: Retrieve a list of Employees and the Projects
each works in, ordered by Dept., and within each Dept.,
alphabetically by Employee last name
SELECT
FROM
WHERE
ORDER BY
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DNAME, LNAME, FNAME, PNAME
DEPARTMENT, EMPLOYEE,
WORKS_ON, PROJECT
DNUMBER=DNO AND SSN=ESSN
AND PNO=PNUMBER
DNAME, LNAME
Default is Ascending - Can be ASC/DESC as we’ll see
in a Later Example
Chapter 5-21
Let’s Jump to In Class Exercises
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Chapter 5-22
SQL Support for Views
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Views are Part of the SQL DDL
Abstracting from Conceptual to External Schema
 View Hides the Details
 One or More Tables in Conceptual Schema May be
Combined (in Part) to Form a View
 Don’t Include FKs and Other Internal Attributes
 Typically, View is Formed by Join of Two or More
Relations Utilizing FKs, PKs, etc.
 As a Result - View is Independent
Once
Formed - View Static/Unchangeable to Insulate
User Applications from Conceptual Schema
Similar in Concept/Intent to “Public Interface”
Chapter 5-23
SQL View Definition
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Features of Views
 View Represents a Restricted Portion (Rows,
Columns) of a Relation - External Schema in SQL
 View is Virtual Table View (Not Stored) and Must
be Re-evaluated Every Time - Dynamic
 Like Relation, a View Can Be Deleted at Any Time
 Attributes Can Be Renamed in View
CREATE VIEW PQ(P#, SUMQTY)
AS SELECT P#, SUM(QTY)
FROM SP
GROUP BY P#;
Reasons for Views
 Security
 Increasing Application-Data Independence
Chapter 5-24
View Definition in Ongoing Example
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First View: Attribute Names are Inherited
CREATE VIEW WORKS_ON1 AS
SELECT FNAME, LNAME, PNAME, HOURS
FROM EMPLOYEE, PROJECT, WORKS_ON
WHERESSN=ESSN AND PNO=PNUMBER ;
Second View: View attribute names are Aliased via a
one-to-one Correspondence with the SELECT-clause
CREATE VIEW DEPT_INFO
(DEPT_NAME, NO_OF_EMPS, TOTAL_SAL) AS
SELECT
DNAME, COUNT (*), SUM (SALARY)
FROM
DEPARTMENT, EMPLOYEE
WHERE
DNUMBER=DNO
GROUP BY
DNAME ;
Chapter 5-25
Queries on Views
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Retrieve the Last Name and First Name of All
Employees Who Work on 'ProjectX'.
SELECT PNAME, FNAME, LNAME
FROM
WORKS_ON1
WHERE PNAME='ProjectX' ;
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Without the View WORKS_ON1, this Query
Specification Would Require Two Join Conditions
A View Can Be Defined to Simplify Frequently
Occurring Queries
DBMS Keeps the View Up-to-date if the Base Tables
on Which the View is Defined are Modified
Hence, the View is Realized at the Time we Specify a
Query on the View
Chapter 5-26
What is View Update Problem?
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Retrieval over View Mirrors a Retrieval over Relation
However, Update over View may cause Problems!
In general, a View Update may Introduce Ambiguity
when there is more than one way to Update Underlying
Relations
Consider the view PQ Created Below:
CREATE VIEW PQ(P#, SUMQTY)
AS SELECT P#, SUM(QTY)
FROM SP
GROUP BY P#;
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Try to Change the Total Quantity SUMQTY of P1 in
PQ from “30” to “40”
Why Does a view Update Problem occur?
Chapter 5-27
The Book Example - DDL - Create Tables
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Create a Table in the MY_BOOK_DB Schema:
What do ISBNNUMBER and PUBLISHERID
Represent?
CREATE TABLE BOOK_CATALOG
(ISBN
ISBNNUMBER NOT NULL,
TITLE
VARCHAR(25),
AUTHORS VARCH(100),
...
PUBLISHER PUBLISHERID,
PRIMARY KEY (ISBN)
FOREIGN KEY (PUBLISHER)
REFERENCES PUBLISHING_HOUSE(NAME);
Chapter 5-28
DDL - Create Tables (continued)
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CREATE TABLE PUBLISHING_HOUSE
(PUB_ID
PUBLISHERID
NOT NULL,
PUB_NAME
VARCHAR(50)
LOC
CITYNAME
CONTACT
ADDRESS,
UNIQUE(PUB_NAME, LOC);
CREATE TABLE ORDER
(ISBN
ISBNNUMBER NOT NULL,
PUBLISHER
PUBLISHERID
DATE
DATE
NOT NULL,
PRIMARY KEY(ISBN, PUBLISHER),
FOREIGN KEY ISBN
REFERENCES BOOK_CATALOG(ISBN),
FOREIGN KEY PUBLISHER
REFERENCES PUBLISHING_HOUSE(PUB_ID));
Chapter 5-29
DDL - Change Table Structure
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Add a Column to a Table:
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ALTER TABLE BOOK
ADD PRICE DECIMAL(7,2),
ADD MEMBER_DISCOUNT, DEFAULT 5;
No DEFAULT Implies NULL Values for all Tuples
Drop a Column from a table
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ALTER TABLE BOOK
DROP PRICE RESTRICT (or CASCADE);
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Restrict: Drop Operation Fails if Column is
Referenced
Cascade: Drop Operation Removes Referencing View
and Constraint Definitions
Chapter 5-30
Views Basis of APIs
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WSDL, SOAP, REST, etc.
View Presents a Limited Picture of the DB
 Defines Data Available to Different
User
Groups
Applications (web or mobile)
Other Systems (for Interoperability/Sharing)
Most Typically Read-Based Views
Update-Based Views Carefully Define
 Insure that Updates Don’t Alter Consistency
 May Limit What
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Chapter 5-31
Views in Medical Domain
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Prescription Object has Multiple Views
MD/Prescriber
 Ability to Write the Script with Drug, Dosage,
Instructions, etc.
Pharmacist -Fill Prescription (Drug dispensed)
 Ability to Substitute Generic for Brand
 A Refill Reduces Future Availability
Patient
 Submit Script to Pharmacy/Insurance
Insurer
 See Drug Dispensed, Script, Approve Payment
Chapter 5-32
Other SQL Query Examples
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Homework 2 Spring 2015
Book Database
Employee/Project/Department Database
Chinook
Northwind
Chapter 5-33
Another Set of Examples- Book DB
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BOOK(ISBN, TITLE,AUTHORS,PRICE,PUBLISHER,YEAR)
ORDER(ISBN, CUST_NAME, LOC, DATE,WEEKDAY)
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Find the books Written by Maier
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SELECT TITLE, ISBN
FROM BOOK;
WHERE AUTHOR LIKE ‘%MAIER’ ;
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Find ISBN of the Books Whose Price is at Least 5%
less than the Average Price of the Books by Maier
SELECT ISBN, PRICE
FROM BOOK
WHERE PRICE*(1-0.05) <
SELECT AVG(PRICE)
FROM BOOK
WHERE AUTHORS LIKE “%Maier”;
Chapter 5-34
Other SQL Search: String Matching
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Wildcard:
%: Matches any Substring
_: Matches any Character
SELECT *
FROM BOOK
WHERE PUBLISHER LIKE “%IEEE%”;
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“%can%” Matches American, Canada, Scandinavian, ...
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“Ca%” Matches Canada, Canadian, ...
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“ %” Matches any string with at least two characters
Chapter 5-35
Other SQL Queries
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Find ISBN and Price of Books Published by ACM
SELECT ISBN, PRICE
FROM BOOK
WHERE PUBLISHER=“ACM”;
Find ISBN and price for all books ordered from Atlanta
with a price over $50
SELECT ISBN, PRICE
FROM BOOK, ORDER
WHERE BOOK.ISBN = ORDER.ISBN
AND ORDER.LOC=‘ATL’
AND PRICE > 50.00;
Note the Distinguishing Between Attributes with Same
Name in Different Tables (TableName.AttributeName)
Chapter 5-36
Using Tuple Variables
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Tuple Variables Simplify Query Since Don’t Need to
Repeat the Entire Table Name
Find ISBN and Price for all Books ordered from
Atlanta with a price over $50
SELECT B.ISBN, B.PRICE
FROM BOOK B, ORDER O
WHERE B.ISBN = O.ISBN AND O.LOC=‘ATL’
AND B.PRICE > 50.00;
Also Useful if Relation is Used “twice” in a Query:
SELECT B1.ISBN, B1.TITLE, B1.AUTHORS
FROM BOOK B1, BOOK B2
WHERE B1.PRICE > B2.PRICE
AND B2.ISBN = “111001100”;
Chapter 5-37
Ordering Results
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Order by Clause Sorts the rows in a Query Result in
Ascending (asc) or Descending (desc) Order
Find all books Published by ACM in the Ascending
order of Price and Descending order of year
SELECT *
FROM BOOK
WHERE PUBLISHER LIKE “ACM%”
ORDER BY PRICE ASC, YEAR DESC;
Questions:
 What Does “*” Indicate?
 What Does ACM% Retrieve?
Chapter 5-38
Set Operations
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Find books written by Mary or Lisa
SELECT *
FROM BOOK
WHERE AUTHORS LIKE “LISA%”
UNION
SELECT *
FROM BOOK
WHERE AUTHORS LIKE “MARY%”;
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SELECT *
FROM BOOK
WHERE AUTHORS LIKE “LISA%”
OR AUTHORS LIKE “MARY%”;
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UNION, INTERSECT, EXCEPT
UNION ALL, INTERSECT ALL, and EXCEPT
ALL Preserve Duplicates
Chapter 5-39
Built-in Aggregate Functions
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Count (COUNT), Sum (SUM), Average (AVG),
Minimum (MIN), Maximum (MAX)
Count Books Ordered on 2/16
SELECT COUNT( *)
FROM ORDER
WHERE ORDER.DATE = “2/16/2000”;
Find the Average Price of Books by each Publisher
SELECT PUBLISHER, AVG(PRICE)
FROM BOOK
GROUP BY PUBLISHER;
Chapter 5-40
Built-in Aggregate Functions
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Find the Average Book Price of all Publishers that have
Published more than 1000 Books
SELECT PUBLISHER, AVG(PRICE)
FROM BOOK
GROUP BY PUBLISHER
HAVING COUNT (ISBN) >= 1000;
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Find the Highest Priced book(s) by Maier
SELECT ISBN, MAX(PRICE)
FROM BOOK
WHERE AUTHORS LIKE “%Maier%”;
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Chapter 5-41
Nested Subqueries
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Nested Subqueries Allow us to Ask More Complex
Questions Regarding the Database Content
Queries are Nested and Involve Set Relationships
Relationships Supported Include:
 Set Membership: IN, NOT IN
 Set Comparison
(=, <, <=, >, >=, <>) ALL
(=, <, <=, >, >=, <>) SOME
 Test Empty Relation: EXISTS, NOT EXISTS
Let’s see Some Examples…
Chapter 5-42
Set Membership: IN, NOT IN
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Find Title of the Books Ordered on Mondays
SELECT DISTINCT TITLE
FROM BOOK
WHERE ISBN IN
(SELECT ISBN
FROM ORDER
WHERE WEEKDAY = “MON”);
Find Titles of Books ordered on Wednesday to Friday
SELECT DISTINCT ISBN
FROM BOOK
WHERE WEEKDAY NOT IN (“MON”, “TUE”);
Chapter 5-43
Set Comparison
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Operators
(=, <, <=, >, >=, <>) ALL
(=, <, <=, >, >=, <>) SOME
Find ISBN of Books Published by ACM, which are
Cheaper than all Books Ordered by Smith
SELECT ISBN
FROM BOOK
WHERE PUBLISHER LIKE “%ACM%”
AND PRICE <
ALL (SELECT B.PRICE
FROM BOOK B, ORDER O
WHERE CUST_NAME LIKE “%SMITH%”
AND B.ISBN = O.ISBN);
Chapter 5-44
EXISTS, NOT EXISTS
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Find ISBN of Books with a Price so low That There is
not any Cheaper Books from ACM
SELECT B.ISBN
FROM BOOK B
WHERE B.PUBLISHER LIKE “%ACM%”
AND
NOT EXISTS
(SELECT T.ISBN
FROM BOOK T
WHERE T.PUBLISHER LIKE “%ACM%”
AND T.PRICE < B.PRICE);
Chapter 5-45
Delete and Update
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Cancel all orders by Mary on 2/17/2000
DELETE FROM BOOK
WHERE DATE=2000-02-17
AND ISBN IN
(SELECT ISBN
FROM ORDER
WHERE CUST_NAME LIKE “%Mary%”);
Update all Orders for Customers on 2/15/2002 by
giving a discount of 5%
UPDATE ORDER
SET PRICE = PRICE * (1-0.05)
WHERE DATE=2002-02-50;
Chapter 5-46
View Concepts/Examples
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REM updatable view
CREATE VIEW LOW-PRICE-BOOKS
AS SELECT *
FROM BOOKS
WHERE PRICE<50.00;
REM moves row outside view
UPDATE LOW-PRICE-BOOKS
SET PRICE = 60.00
WHERE PUBLISHER LIKE “%ACM%”;
REMcreate row outside view
INSERT INTO LOW-PRICE-BOOKS
VALUES (“1010110022”, ”Java Beans”,
“Smith”, 45, ”ACM”);
REM prevents updates outside the view
CREATE VIEW LOW-PRICE-BOOKS
AS SELECT *
FROM BOOK
WHERE PRICE<50.00
WITH CHECK OPTION;
Chapter 5-47
Homework 3 from Spr 2015
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Problem 6.18 from the 6th
edition done in SQL and
NOT relational expressions
Chapter 5-48
Problem 6.18 in 6th edition
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a. How many copies of the book titled The Lost Tribe are owned by the
library branch whose name is ‘Sharpstown’?
SELECT NoOfCopies
FROM ( (BOOK NATURAL JOIN BOOK_COPIES )
NATURAL JOIN LIBRARY_BRANCH )
WHERE Title='The Lost Tribe' AND BranchName='Sharpstown’
BaB= (BOOKCOPIES * (Title=‘The Lost Tribe’ (BOOK))) )
BookId
Ans = No_Of_Copies( (BranchName=‘Sharpstown’ (LIBRARY-BRANCH)) * BaB)
BranchID
Chapter 5-49
Problem 6.18 in 6th edition
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b. How many copies of the book titled The Lost Tribe are owned by each
library branch?
SELECT BranchName, NoOfCopies
FROM ( (BOOK NATURAL JOIN BOOK_COPIES )
NATURAL JOIN LIBRARY_BRANCH )
WHERE Title='The Lost Tribe'
CaB = BOOKCOPIES * LIBRARY_BRANCH)
BranchId
Ans = BranchName, No_Of_Copies( (Title=‘The Lost Tribe’ (BOOK)) * CaB)
BookId
Chapter 5-50
Problem 6.18 in 6th edition
c. Retrieve the names of all borrowers who do not have any books checked out
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SELECT Name
FROM BORROWER B
WHERE NOT EXIST
( SELECT *
FROM BOOK_LOANS L
WHERE B.CardNo = L.CardNo )
d. For each book that is loaned out from the Sharpstown branch and whose
Due_date is today, retrieve the book title, the borrower’s name, and the
borrower’s address.
SELECT B.Title, R.Name, R.Address
FROM BOOK B, BORROWER R, BOOK_LOANS BL, LIBRARY_BRANCH LB
WHERE LB.BranchName='Sharpstown' AND LB.BranchId=BL.BranchId AND
BL.DueDate='today' AND BL.CardNo=R.CardNo AND BL.BookId=B.BookId
Chapter 5-51
Problem 8.11/6.18 in 6th edtion
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e. For each library branch, retrieve the branch name and the total number
of books loaned out from that branch.
SELECT L.BranchName, COUNT(*)
FROM BOOK_COPIES B, LIBRARY_BRANCH L
WHERE B.BranchId = L.BranchId
GROUP BY L.BranchName
f. Retrieve the names, addresses, and number of books checked out for all
borrowers who have more than five books checked out.
SELECT B.CardNo, B.Name, B.Address, COUNT(*)
FROM BORROWER B, BOOK_LOANS L
WHERE B.CardNo = L.CardNo
GROUP BY B.CardNo
HAVING COUNT(*) > 5
g. For each book authored (or coauthored) by Stephen King, retrieve the
title and the number of copies owned by the library branch whose name
is Central.
SELECT TItle, NoOfCopies
FROM ( ( (BOOK_AUTHORS NATURAL JOIN BOOK)
NATURAL JOIN BOOK_COPIES)
NATURAL JOIN LIBRARY_BRANCH)
WHERE Author_Name = 'Stephen King' and BranchName = 'Central'
Chapter 5-52
Homework 3 from Spr 2015

Problem 4.10 in 6th edition
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Chapter 5-53
Problem 4.10 in 6th edition
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Chapter 5-54
Problem 4.10 in 6th edition
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(a) Retrieve the names of employees in department 5 who
work more than 10 hours per week on the 'ProductX' project.
SELECT LNAME, FNAME
FROM EMPLOYEE, WORKS_ON, PROJECT
WHERE DNO=5 AND SSN=ESSN AND
PNO=PNUMBER AND PNAME='ProductX' AND HOURS>10
SELECT LNAME, FNAME
FROM EMPLOYEE
WHERE DNO=5 AND SSN IN
( SELECT ESSN
FROM WORKS_ON
WHERE HOURS>10 AND PNO IN
( SELECT PNUMBER
FROM PROJECT
WHERE PNAME='ProductX' ) )
LNAME
FNAME
Smith
John
English Joyce
Chapter 5-55
Problem 4.10 in 6th edition
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(b) List the names of employees who have a dependent
with the same first name as themselves.
SELECT LNAME, FNAME
FROM EMPLOYEE, DEPENDENT
WHERE SSN=ESSN AND FNAME=DEPENDENT_NAME
Another possible SQL query uses nesting as follows:
SELECT LNAME, FNAME
FROM EMPLOYEE
WHERE EXISTS
( SELECT *
FROM DEPENDENT
WHERE FNAME=DEPENDENT_NAME
AND SSN=ESSN )
Result (empty):
Chapter 5-56
Problem 4.10 in 6th edition
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(c) Find the names of employees that are
directly supervised by 'Franklin Wong'.
LNAME FNAME
Smith John
Narayan Ramesh
English Joyce
SELECT E.LNAME, E.FNAME
FROM EMPLOYEE E, EMPLOYEE S
WHERE S.FNAME='Franklin' AND
S.LNAME='Wong' AND E.SUPERSSN=S.SSN
Another possible SQL query uses nesting as follows:
SELECT LNAME, FNAME
FROM EMPLOYEE
WHERE SUPERSSN IN
( SELECT SSN
FROM EMPLOYEE
WHERE FNAME='Franklin' AND LNAME='Wong' )
Chapter 5-57
Problem 4.10 in 6th edition
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(d) For each project, list the project name and the total hours per week (by
all employees) spent on that project.
SELECT PNAME, SUM (HOURS)
FROM PROJECT, WORKS_ON
WHERE PNUMBER=PNO
GROUP BY PNAME
Result:
PNAME
SUM(HOURS)
ProductX
52.5
ProductY
37.5
ProductZ
50.0
Computerization55.0
Reorganization
25.0
Newbenefits
55.0
Chapter 5-58
Problem 4.10 in 6th edition
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(e) Retrieve the names of employees who work on every project.
SELECT LNAME, FNAME
FROM EMPLOYEE
WHERE NOT EXISTS
( SELECT PNUMBER
FROM PROJECT
WHERE NOT EXISTS
( SELECT *
FROM WORKS_ON
WHERE PNUMBER=PNO AND ESSN=SSN ) )
Result (empty):
Chapter 5-59
Problem 4.10 in 6th edition
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(f) Retrieve the names of employees who do not
work on any project.
SELECT LNAME, FNAME
FROM EMPLOYEE
WHERE NOT EXISTS
( SELECT *
FROM WORKS_ON
WHERE ESSN=SSN )
Result (empty):
Chapter 5-60
Problem 4.10 in 6th edition
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(g) For each department, retrieve the department name,
and the average salary of employees working in that department.
SELECT DNAME, AVG (SALARY)
FROM DEPARTMENT, EMPLOYEE
WHERE DNUMBER=DNO
GROUP BY DNAME
Result:
DNAME
AVG(SALARY)
Research
33250
Administration 31000
Headquarters
55000
Chapter 5-61
Problem 4.10 in 6th edition
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(i) Find the names and addresses of employees who work on at least one
project located in Houston but whose department has no location in Houston.
SELECT LNAME, FNAME, ADDRESS
FROM EMPLOYEE
WHERE EXISTS
( SELECT *
FROM WORKS_ON, PROJECT
WHERE SSN=ESSN AND PNO=PNUMBER
AND PLOCATION='Houston' )
AND NOT EXISTS
( SELECT *
FROM DEPT_LOCATIONS
WHERE DNO=DNUMBER
AND DLOCATION='Houston' )
Result:
LNAME
FNAME
ADDRESS
Wallace Jennifer 291 Berry, Bellaire, TX
Chapter 5-62
Problem 4.10 in 6th edition
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(j) List the last names of department managers who have no
dependents.
SELECT LNAME, FNAME
FROM EMPLOYEE
WHERE EXISTS
( SELECT *
FROM DEPARTMENT
WHERE SSN=MGRSSN )
AND NOT EXISTS
( SELECT *
FROM DEPENDENT
WHERE SSN=ESSN )
Result:
LNAME FNAME
Borg James
Chapter 5-63
Homework 4 from Spring 2015
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Problem 3.3 for the Chinook Database Schema
a.
Find the list of all Customers and Employees that have the same
last name and print out the Last Name (only once), Address,
City, and State for each.
b. Find and print the Customer Names (First and Last) and
Company Name of all Customers that have purchased a Rock
Album. (where the Name is ‘Rock’ in the Genre table).
c.
Find all of print album name and tracks of all of the albums by
the composer James Hetfield, grouped by Album.
d. For each customer that lives in Canada (CA- Country attribute of
Customer), find all invoices and for each result, and print
Customer Last Name, number of invoices for customer, and the
total amount paid for all invoices.
Chapter 5-64
Chinook EER
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Chapter 5-65
Problems 3.3a and 3.3b

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Find the list of all Customers and Employees that have the same last name
and print out the Last Name (only once), Address, City, and State for each.
SELECT employee.LastName, employee.Address, employee.City,
employee.State, customer.Address, customer.city, customer.state
FROM chinook.employee, chinook.customer
WHERE employee.LastName = customer.LastName;

Find and print the Customer Names (First and Last) and Company Name of
all Customers that have purchased a Rock Album. (where the Name is ‘Rock’
in the Genre table).
SELECT DISTINCT customer.FirstName, customer.LastName, customer.Company
FROM chinook.customer, chinook.invoice, chinook.invoiceline, chinook.track,
chinook.genre
WHERE customer.CustomerId = invoice.CustomerId
AND invoice.InvoiceId = invoiceline.InvoiceId AND invoiceline.TrackId =
track.TrackId
Chapter 5-66
Problems 3.3c and 3.d

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Find all of print album name and tracks of all of the albums by the composer
James Hetfield, grouped by Album.
SELECT album.title, track.name
FROM chinook.album, chinook.track
WHERE album.AlbumId = track.AlbumId AND track.Composer LIKE
'%Hetfield%';

For each customer that lives in Canada (CA- Country attribute of Customer),
find all invoices and for each result, and print Customer Last Name, number
of invoices for customer, and the total amount paid for all invoices.

SELECT customer.LastName, COUNT(invoice.InvoiceId),
SUM(invoice.total)
FROM chinook.customer, chinook.invoice
WHERE customer.Country = 'Canada' AND customer.CustomerId =
invoice.CustomerId
Chapter 5-67
Homework 4 from Spring 2015
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Problem 3.4 for the Northwind Database Schema
a.
Find and print the company names and company addresses of all
Suppliers that supply the category name Seafood.
b. Count and print the number of suppliers for each of the eight
different categories of food which by name are: Beverages,
Condiments, Confections, Dairy Products, Grains/Cereals,
Meat/Poultry, Produce, Seafood.
c.
For each country (ShipCountry in Orders), total the Freight
charges. The countries are: France, Germany, Brazil, Belgium,
Switzerland, Venezuela, Austria, Mexico, USA, Sweden,
Finland, Italy, Spain, UK, Ireland, Portugal, Canada, Denmark,
Poland, Norway, Argentina
Chapter 5-68
Northwind Schema
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Northwind
EER
Chapter 5-69
Problem 4.3a

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Find and print the company names and company addresses of all
Suppliers that supply the category name Seafood.
SELECT DISTINCT suppliers.CompanyName, suppliers.Address
FROM northwind.suppliers, northwind.categories, northwind.products
WHERE suppliers.SupplierID = products.SupplierID
AND categories.CategoryID = products.CategoryID
AND categories.CategoryName = 'Seafood';
Chapter 5-70
Problem 4.3b

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Count and print the number of suppliers for each of the eight different
categories of food which by name are: Beverages, Condiments, Confections,
Dairy Products, Grains/Cereals, Meat/Poultry, Produce, Seafood.
SELECT categories.CategoryName, COUNT(suppliers.SupplierID)
FROM northwind.categories, northwind.products, northwind.suppliers
WHERE suppliers.SupplierID = products.SupplierID
AND products.categoryID = categories.CategoryID
GROUP BY categories.CategoryName;
Chapter 5-71
Problem 4.3c

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For each country (ShipCountry in Orders), total the Freight
charges. The countries are: France, Germany, Brazil, Belgium,
Switzerland, Venezuela, Austria, Mexico, USA, Sweden,
Finland, Italy, Spain, UK, Ireland, Portugal, Canada, Denmark,
Poland, Norway, Argentina
SELECT orders.ShipCountry, SUM(orders.Freight)
FROM northwind.orders
GROUP BY orders.ShipCountry;
Chapter 5-72
Concluding Remarks

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What have we Seen in Chapter 5?
 Complex Data Manipulation in SQL
Nested
Queries
Grouping and Aggregation
Order by and Having
Views
Sets
Tuple Variable
Other Complex Queries and Operations

Strongly Encouraged to Engage in Practice with your
DBMS of Choice for your Project
Chapter 5-73