Chapter 8 Advanced SQL
Chapter 8
Advanced SQL
NOTE
Several points are worth emphasizing:
 We have provided the SQL scripts for the chapters. These scripts are intended to facilitate the
flow of the material presented to the class. However, given the comments made by our
students, the scripts should not replace the manual typing of the SQL commands by students.
Some students learn SQL better when they have a chance to type their own commands and get
the feedback provided by their errors. We recommend that the students use their lab time to
practice the commands manually.
 In this chapter, most of the queries are executed in the Oracle RDBMS. This chapter uses
features that are restricted to client/server DBMSes. Such features are not commonly available
in a desktop DBMS such as Microsoft Access.
Answers to Review Questions
1. The relational set operators UNION, INTERSECT, and MINUS work properly only if the
relations are union-compatible. What does union-compatible mean, and how would you check
for this condition?
Union compatible means that the relations yield attributes with identical names and compatible data
types. That is, the relation A(c1,c2,c3) and the relation B(c1,c2,c3) have union compatibility if the
columns have the same names, are in the same order, and the columns have “compatible” data types.
Compatible data types do not require that the attributes be exactly identical -- only that they are
comparable. For example, VARCHAR(15) and CHAR(15) are comparable, as are NUMBER (3,0)
and INTEGER, and so on.
2. What is the difference between UNION and UNION ALL? Write the syntax for each.
UNION yields unique rows. In other words, UNION eliminates duplicates rows. On the other hand,
a UNION ALL operator will yield all rows of both relations, including duplicates. Notice that for
two rows to be duplicated, they must have the same values in all columns.
To illustrate the difference between UNION and UNION ALL, let’s assume two relations:
A (ID, Name) with rows (1, Lake, 2, River, and 3, Ocean)
and
B (ID, Name) with rows (1, River, 2, Lake, and 3, Ocean).
Given this description,
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SELECT * FROM A
UNION
SELECT * FROM B
will yield:
ID
1
2
3
1
2
Name
Lake
River
Ocean
River
Lake
while
SELECT * FROM A
UNION ALL
SELECT * FROM B
will yield:
ID
1
2
3
1
2
3
Name
Lake
River
Ocean
River
Lake
Ocean
3. Suppose that you have two tables, EMPLOYEE and EMPLOYEE_1. The EMPLOYEE table
contains the records for three employees: Alice Cordoza, John Cretchakov, and Anne
McDonald. The EMPLOYEE_1 table contains the records for employees John Cretchakov and
Mary Chen. Given that information, what is the query output for the UNION query? (List the
query output.)
The query output will be:
Alice Cordoza
John Cretchakov
Anne McDonald
Mary Chen
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4. Given the employee information in Question 3, what is the query output for the UNION ALL
query? (List the query output.)
The query output will be:
Alice Cordoza
John Cretchakov
Anne McDonald
John Cretchakov
Mary Chen
5. Given the employee information in Question 3, what is the query output for the INTERSECT
query? (List the query output.)
The query output will be:
John Cretchakov
6. Given the employee information in Question 3, what is the query output for the MINUS
query? (List the query output.)
This question can yield two different answers. If you use
SELECT * FROM EMPLOYEE
MINUS
SELECT * FROM EMPLOYEE_1
the answer is
Alice Cordoza
Ann McDonald
If you use
SELECT * FROM EMPLOYEE_1
MINUS
SELECT * FROM EMPLOYEE
the answer is
Mary Chen
7. What is a CROSS JOIN? Give an example of its syntax.
A CROSS JOIN is identical to the PRODUCT relational operator. The CROSS JOIN is also known
as the Cartesian product of two tables. For example, if you have two tables, AGENT, with 10 rows
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and CUSTOMER, with 21 rows, the CROSS JOIN resulting set will have 210 rows and will include
all of the columns from both tables. Syntax examples are:
SELECT * FROM CUSTOMER CROSS JOIN AGENT;
or
SELECT * FROM CUSTOMER, AGENT
If you do not specify a join condition when joining tables, the result will be a CROSS Join or
PRODUCT operation.
8. What three join types are included in the OUTER JOIN classification?
An OUTER JOIN is a type of JOIN operation that yields all rows with matching values in the join
columns as well as all unmatched rows. (Unmatched rows are those without matching values in the
join columns). The SQL standard prescribes three different types of join operations:
LEFT [OUTER] JOIN
RIGHT [OUTER] JOIN
FULL [OUTER] JOIN.
The LEFT [OUTER] JOIN will yield all rows with matching values in the join columns, plus all of
the unmatched rows from the left table. (The left table is the first table named in the FROM clause.)
The RIGHT [OUTER] JOIN will yield all rows with matching values in the join columns, plus all of
the unmatched rows from the right table. (The right table is the second table named in the FROM
clause.)
The FULL [OUTER] JOIN will yield all rows with matching values in the join columns, plus all the
unmatched rows from both tables named in the FROM clause.
9. Using tables named T1 and T2, write a query example for each of the three join types you
described in Question 8. Assume that T1 and T2 share a common column named C1.
LEFT OUTER JOIN example:
SELECT * FROM T1 LEFT OUTER JOIN T2 ON T1.C1 = T2.C1;
RIGHT OUTER JOIN example:
SELECT * FROM T1 RIGHT OUTER JOIN T2 ON T1.C1 = T2.C1;
FULL OUTER JOIN example:
SELECT * FROM T1 FULL OUTER JOIN T2 ON T1.C1 = T2.C1;
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10. What is a subquery, and what are its basic characteristics?
A subquery is a query (expressed as a SELECT statement) that is located inside another query. The
first SQL statement is known as the outer query, the second is known as the inner query or subquery.
The inner query or subquery is normally executed first. The output of the inner query is used as the
input for the outer query. A subquery is normally expressed inside parenthesis and can return zero,
one, or more rows and each row can have one or more columns.
A subquery can appear in many places in a SQL statement:
 as part of a FROM clause,
 to the right of a WHERE conditional expression,
 to the right of the IN clause,
 in a EXISTS operator,
 to the right of a HAVING clause conditional operator,
 in the attribute list of a SELECT clause.
Examples of subqueries are:
INSERT INTO PRODUCT
SELECT * FROM P;
DELETE FROM PRODUCT
WHERE V_CODE IN (SELECT V_CODE FROM VENDOR
WHERE V_AREACODE = ‘615’);
SELECT
FROM
WHERE
V_CODE, V_NAME
VENDOR
V_CODE NOT IN (SELECT V_CODE FROM PRODUCT);
11. What is a correlated subquery? Give an example.
A correlated subquery is subquery that executes once for each row in the outer query. This process is
similar to the typical nested loop in a programming language. Contrast this type of subquery to the
typical subquery that will execute the innermost subquery first, and then the next outer query …
until the last outer query is executed. That is, the typical subquery will execute in serial order, one
after another, starting with the innermost subquery. In contrast, a correlated subquery will run the
outer query first, and then it will run the inner subquery once for each row returned in the outer
subquery.
For example, the following subquery will list all the product line sales in which the “units sold”
value is greater than the “average units sold” value for that product (as opposed to the average for all
products.)
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SELECT
FROM
WHERE
INV_NUMBER, P_CODE, LINE_UNITS
LINE LS
LS.LINE_UNITS > (SELECT
AVG(LINE_UNITS) FROM LINE LA
WHERE
LA.P_CODE = LS.P_CODE);
The previous nested query will execute the inner subquery once to compute the average sold units
for each product code returned by the outer query.
12. What MS Access/SQL Server function should you use to calculate the number of days between
the current date and January 25, 1999?
SELECT DATE()-#25-JAN-1999#
NOTE: In MS Access you do not need to specify a FROM clause for this type of query.
13. What Oracle function should you use to calculate the number of days between the current date
and January 25, 1999?
SELECT
FROM
SYSDATE – TO_DATE(’25-JAN-1999’, ‘DD-MON-YYYY’)
DUAL;
Note that in Oracle, the SQL statement requires the use of the FROM clause. In this case, you must
use a DUAL table. (A DUAL table is a dummy “virtual” table provided by Oracle for this type of
query.
14. Suppose that a PRODUCT table contains two attributes, PROD_CODE and VEND_CODE.
Those two attributes have values of ABC, 125, DEF, 124, GHI, 124, and JKL, 123, respectively.
The VENDOR table contains a single attribute, VEND_CODE, with values 123, 124, 125, and
126, respectively. (The VEND_CODE attribute in the PRODUCT table is a foreign key to the
VEND_CODE in the VENDOR table.) Given that information, what would be the query
output for:
Because the common attribute is V_CODE, the output will only show the V_CODE values
generated by the each query.
a. A UNION query based on these two tables?
125,124,123,126
b. A UNION ALL query based on these two tables?
125,124,124,123,123,124,125,126
c. An INTERSECT query based on these two tables?
123,124,125
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d. A MINUS query based on these two tables?
If you use PRODUCT MINUS VENDOR, the output will be NULL
If you use VENDOR MINUS PRODUCT, the output will be 126
15. What Oracle string function should you use to list the first three characters of a company’s
EMP_LNAME values? Give an example, using a table named EMPLOYEE.
You must use the SUBSTR function as illustrated next:
SELECT SUBSTR(EMP_LNAME,1,3) FROM EMPLOYEE;
16. What is an Oracle sequence? Write its syntax.
An Oracle sequence is a special type of object that generates unique numeric values in ascending or
descending order. You can use a sequence to assign values to a primary key field in a table.
A sequence provides functionality similar to the Autonumber data type in MS Access. For example,
both, sequences and Autonumber data types provide unique ascending or descending values.
However, there are some subtle differences between the two:
 In MS Access an Autonumber is a data type; in Oracle a sequence is a completely
independent object, rather than a data type.
 In MS Access, you can only have one Autonumber per table; in Oracle you can have as many
sequences as you want and they are not tied to any particular table.
 In MS Access, the Autonumber data type is tied to a field in a table; in Oracle, the sequencegenerated value is not tied to any field in any table and can, therefore, be used on any
attribute in any table.
The syntax used to create a sequence is:
CREATE SEQUENCE CUS_NUM_SQ START WITH 100 INCREMENT BY 10 NOCACHE;
17. What is a trigger, and what is its purpose? Give an example.
A trigger is a block of PL/SQL code that is automatically invoked by the DBMS upon the
occurrence of a data manipulation event (INSERT, UPDATE or DELETE.) Triggers are always
associated with a table and are invoked before or after a data row is inserted, updated, or deleted.
Any table can have one or more triggers.
Triggers provide a method of enforcing business rules such as:
 A customer making a credit purchase must have an active account.
 A student taking a class with a prerequisite must have completed that prerequisite with a B
grade.
 To be scheduled for a flight, a pilot must have a valid medical certificate and a valid training
completion record.
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Triggers are also excellent for enforcing data constraints that cannot be directly enforced by the data
model. For example, suppose that you must enforce the following business rule:
If the quantity on hand of a product falls below the minimum quantity, the
P_REORDER attribute must the automatically set to 1.
To enforce this business rule, you can create the following TRG_PRODUCT_REORDER trigger:
CREATE OR REPLACE TRIGGER TRG_PRODUCT_REORDER
BEFORE INSERT OR UPDATE OF P_ONHAND, P_MIN ON PRODUCT
FOR EACH ROW
BEGIN
IF :NEW.P_ONHAND <= :NEW.PROD_MIN THEN
NEW.P_REORDER := 1;
ELSE
:NEW.P_REORDER := 0;
END IF;
END;
18. What is a stored procedure, and why is it particularly useful? Give an example.
A stored procedure is a named block of PL/SQL and SQL statements. One of the major advantages
of stored procedures is that they can be used to encapsulate and represent business transactions. For
example, you can create a stored procedure to represent a product sale, a credit update, or the
addition of a new customer. You can encapsulate SQL statements within a single stored procedure
and execute them as a single transaction.
There are two clear advantages to the use of stored procedures:
1. Stored procedures substantially reduce network traffic and increase performance. Because
the stored procedure is stored at the server, there is no transmission of individual SQL
statements over the network.
2. Stored procedures help reduce code duplication through code isolation and code sharing
(creating unique PL/SQL modules that are called by application programs), thereby
minimizing the chance of errors and the cost of application development and maintenance.
For example, the following PRC_LINE_ADD stored procedure will add a new invoice line to the
LINE table and it will automatically retrieve the correct price from the PRODUCT table.
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CREATE OR REPLACE PROCEDURE PRC_LINE_ADD
(W_LN IN NUMBER, W_P_CODE IN VARCHAR2, W_LU NUMBER)
AS
W_LP NUMBER := 0.00;
BEGIN
-- GET THE PRODUCT PRICE
SELECT P_PRICE INTO W_LP
FROM PRODUCT
WHERE P_CODE = W_P_CODE;
-- ADDS THE NEW LINE ROW
INSERT INTO LINE
VALUES(INV_NUMBER_SEQ.CURRVAL, W_LN, W_P_CODE, W_LU, W_LP);
DBMS_OUTPUT.PUT_LINE('Invoice line ' || W_LN || ' added');
END;
19. What is embedded SQL, and how is it used?
Embedded SQL is a term used to refer to SQL statements that are contained within an application
programming language such as COBOL, C++, ASP, Java, or ColdFusion. The program may be a
standard binary executable in Windows or Linux, or it may be a Web application designed to run
over the Internet. No matter what language you use, if it contains embedded SQL statements it is
called the host language. Embedded SQL is still the most common approach to maintaining
procedural capabilities in DBMS-based applications.
For example, the following embedded SQL code will delete the employee 109, George Smith, from
the EMPLOYEE table:
EXEC SQL
DELETE FROM EMPLOYEE WHERE EMP_NUM = 109;
END-EXEC.
Remember that the preceding embedded SQL statement is compiled to generate an executable
statement. Therefore, the statement is fixed permanently; that is, it cannot change -- unless, of
course, the programmer changes it. Each time the program runs, it deletes the same row. In short, the
preceding code is good only for the first run; all subsequent runs will more than likely give an error.
(Employee 109 is deleted the first time the embedded statement is executed. If you run the code
again, it tries to delete an employee who has already been deleted.) Clearly, this code would be more
useful if you are able to specify a variable to indicate which employee number is to be deleted.
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20. What is dynamic SQL, and how does it differ from static SQL?
Dynamic SQL is a term used to describe an environment in which the SQL statement is not known in
advance; instead, the SQL statement is generated at run time. In a dynamic SQL environment, a
program can generate the SQL statements (at run time) that are required to respond to ad-hoc
queries. In such an environment, neither the programmers nor end users are likely to know precisely
what kind of queries are to be generated or how those queries are to be structured. For example, a
dynamic SQL equivalent of the example shown in question 19 might be:
SELECT :W_ATTRIBUTE_LIST
FROM
:W_TABLE
WHERE :W_CONDITION;
Note that the attribute list and the condition are not known until the end user specifies them.
W_TABLE, W_ATRIBUTE_LIST and W_CONDITION are text variables that contain the end-user
input values used in the query generation. Because the program uses the end user input to build the
text variables, the end user can run the same program multiple times to generate different outputs.
For example, in one case the end user may one to know what products have a price less than $100; in
another case, the end user may want to know how many units of a given product are available for
sale at any given moment.
Although dynamic SQL is clearly flexible, such flexibility carries a price. Dynamic SQL tends to be
much slower that static SQL and it requires more computer resources (overhead). In addition, you
are more likely to find different levels of support and incompatibilities between DBMS vendors.
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Problem Solutions
Use the database tables in Figure P8.1 as the basis for problems 1-18.
ONLINE CONTENT
The Ch08_SimpleCo database is located in the Online Student Companion. This database is
stored in Microsoft Access format. The Student Online Companion also includes SQL script
files (Oracle and SQLServer) for all of the data sets used throughout the book.
Figure P8.1 Ch08_SimpleCo Database Tables
1. Create the tables. (Use the MS Access example shown in Figure P8.1 to see what table names
and attributes to use.)
CREATE TABLE CUSTOMER (
CUST_NUM
NUMBER PRIMARY KEY,
CUST_LNAME
VARCHAR(20),
CUST_FNAME
VARCHAR(20),
CUST_BALANCE
NUMBER);
CREATE TABLE CUSTOMER_2 (
CUST_NUM
NUMBER PRIMARY KEY,
CUST_LNAME
VARCHAR(20),
CUST_FNAME
VARCHAR(20));
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CREATE TABLE INVOICE (
INV_NUM
NUMBER PRIMARY KEY,
CUST_NUM
NUMBER,
INV_DATE
DATE,
INV_AMOUNT
NUMBER);
2. Insert the data into the tables you created in Problem 1.
INSERT INTO CUSTOMER VALUES(1000 ,'Smith' ,'Jeanne' ,1050.11);
INSERT INTO CUSTOMER VALUES(1001 ,'Ortega' ,'Juan' ,840.92);
INSERT INTO CUSTOMER_2 VALUES(2000 ,'McPherson' ,'Anne');
INSERT INTO CUSTOMER_2 VALUES(2001 ,'Ortega' ,'Juan');
INSERT INTO CUSTOMER_2 VALUES(2002 ,'Kowalski' ,'Jan');
INSERT INTO CUSTOMER_2 VALUES(2003 ,'Chen' ,'George');
INSERT INTO INVOICE VALUES(8000 ,1000 ,'23-APR-2008' ,235.89);
INSERT INTO INVOICE VALUES(8001 ,1001 ,'23-MAR-2008' ,312.82);
INSERT INTO INVOICE VALUES(8002 ,1001 ,'30-MAR-2008' ,528.1);
INSERT INTO INVOICE VALUES(8003 ,1000 ,'12-APR-2008' ,194.78);
INSERT INTO INVOICE VALUES(8004 ,1000 ,'23-APR-2008' ,619.44);
3. Write the query that will generate a combined list of customers (from tables CUSTOMER and
CUSTOMER_2) that do not include the duplicate customer records. (Note that only the
customer named Juan Ortega shows up in both customer tables.)
SELECT CUST_LNAME, CUST_FNAME FROM CUSTOMER
UNION
SELECT CUST_LNAME, CUST_FNAME FROM CUSTOMER_2;
4. Write the query that will generate a combined list of customers to include the duplicate
customer records.
SELECT CUST_LNAME, CUST_FNAME FROM CUSTOMER
UNION ALL
SELECT CUST_LNAME, CUST_FNAME FROM CUSTOMER_2;
5. Write the query that will show only the duplicate customer records.
We have shown both Oracle and MS Access query formats:
In Oracle:
SELECT CUST_LNAME, CUST_FNAME FROM CUSTOMER
INTERSECT
SELECT CUST_LNAME, CUST_FNAME FROM CUSTOMER_2;
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In MS Access:
SELECT C.CUST_LNAME, C.CUST_FNAME
FROM CUSTOMER AS C, CUSTOMER_2 AS C2
WHERE C.CUST_LNAME=C2.CUST_LNAME AND C.CUST_FNAME=C2.CUST_FNAME;
Because Access doesn’t support the INTERSECT SQL operator, you need to list only the rows in
which all the attributes match.
6. Write the query that will generate only the records that are unique to the CUSTOMER_2
table.
We have shown both Oracle and MS Access query formats:
In Oracle:
SELECT CUST_LNAME, CUST_FNAME FROM CUSTOMER_2
MINUS
SELECT CUST_LNAME, CUST_FNAME FROM CUSTOMER;
In MS Access:
SELECT
FROM
WHERE
C2.CUST_LNAME, C2.CUST_FNAME
CUSTOMER_2 AS C2
C2.CUST_LNAME + C2.CUST_FNAME NOT IN
(SELECT C1.CUST_LNAME + C1.CUST_FNAME FROM CUSTOMER C1);
Because Access doesn’t support the MINUS SQL operator, you need to list only the rows that are in
CUSTOMER_2 that do not have a matching row in CUSTOMER.
7. Write the query to show the invoice number, the customer number, the customer name, the
invoice date, and the invoice amount for all customers with a customer balance of $1,000 or
more.
This command will run in Oracle and in MS Access:
SELECT INV_NUM, CUSTOMER.CUST_NUM, CUST_LNAME, CUST_FNAME, INV_DATE, INV_AMOUNT
FROM
INVOICE INNER JOIN CUSTOMER ON INVOICE.CUST_NUM=CUSTOMER.CUST_NUM
WHERE CUST_BALANCE>=1000;
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8. Write the query that will show the invoice number, the invoice amount, the average invoice
amount, and the difference between the average invoice amount and the actual invoice
amount.
There are at least two ways to do this query.
SELECT
FROM
GROUP BY
INV_NUM, AVG_INV, (INV_AMOUNT - AVG_INV) AS DIFF
INVOICE, (SELECT AVG(INV_AMOUNT) AS AVG_INV FROM INVOICE)
INV_NUM, AVG_INV, INV_AMOUNT- AVG_INV
Another way to write this query is:
SELECT
FROM
GROUP BY
INV_NUM, INV_AMOUNT,
(SELECT AVG(INV_AMOUNT) FROM INVOICE) AS AVG_INV,
(INV_AMOUNT-(SELECT AVG(INV_AMOUNT) FROM INVOICE)) AS DIFF
INVOICE
INV_NUM, INV_AMOUNT;
The preceding code examples will run in both Oracle and MS Access.
9. Write the query that will write Oracle sequences to produce automatic customer number and
invoice number values. Start the customer numbers at 1000 and the invoice numbers at 5000.
The following code will only run in Oracle:
CREATE SEQUENCE CUST_NUM_SQ START WITH 1000 NOCACHE;
CREATE SEQUENCE INV_NUM_SQ START WITH 5000 NOCACHE;
10. Modify the CUSTOMER table to included two new attributes: CUST_DOB and CUST_AGE.
Customer 1000 was born on March 15, 1979, and customer 1001 was born on December 22,
1988.
In Oracle:
ALTER TABLE CUSTOMER ADD (CUST_DOB DATE) ADD (CUST_AGE NUMBER);
The SQL code required to enter the date values is:
UPDATE CUSTOMER
SET CUST_DOB = ’15-MAR-1979’
WHERE CUST_NUM = 1000;
UPDATE CUSTOMER
SET CUST_DOB = ‘2-DEC-1988’
WHERE CUST_NUM = 1001;
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11. Assuming you completed problem 10, write the query that will list the names and ages of your
customers.
In Oracle:
SELECT CUST_LNAME, CUST_FNAME, ROUND((SYSDATE-CUST_DOB)/365,0) AS AGE
FROM
CUSTOMER;
In MS Access:
SELECT CUST_LNAME, CUST_FNAME, ROUND((DATE()-CUST_DOB)/365,0) AS AGE
FROM
CUSTOMER;
NOTE
The correct age computation may be computed by
INT((DATE()-CUST_DOB)/365)
However, students have not (yet) seen the INT function at this point -- which is why we used
ROUND() function.
12. Assuming the CUSTOMER table contains a CUST_AGE attribute, write the query to update
the values in that attribute. Hint: Use the results of the previous query.
In Oracle:
UPDATE CUSTOMER
SET CUST_AGE = ROUND((SYSDATE-CUST_DOB)/365,0);
In MS Access:
UPDATE CUSTOMER
SET CUST_AGE = ROUND((DATE()-CUST_DOB)/365,0);
13. Write the query that will list the average age of your customers. (Assume that the
CUSTOMER table has been modified to include the CUST_DOB and the derived CUST_AGE
attribute.)
SELECT AVG(CUST_AGE) FROM CUSTOMER;
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14. Write the trigger to update the CUST_BALANCE in the CUSTOMER table when a new
invoice record is entered. (Assume that the sale is a credit sale.) Test the trigger using the
following new INVOICE record:
8005, 1001, ’27-APR-08’, 225.40
Name the trigger trg_updatecustbalance.
CREATE OR REPLACE TRIGGER TRG_UPDATECUSTBALANCE
AFTER INSERT ON INVOICE
FOR EACH ROW
BEGIN
UPDATE
CUSTOMER
SET
CUST_BALANCE = CUST_BALANCE + :NEW.INV_AMOUNT
WHERE CUST_NUM = :NEW.CUST_NUM;
END;
To test the trigger you do the following:
SELECT * FROM CUSTOMER;
INSERT INTO INVOICE VALUES (8005,1001,’27-APR-08’,225.40);
SELECT * FROM CUSTOMER;
15. Write a procedure to add a new customer to the CUSTOMER table. Use the following values
in the new record:
1002, ‘Rauthor’, ‘Peter’, 0.00
Name the procedure prc_cust_add. Run a query to see if the record has been added.
CREATE OR REPLACE PROCEDURE PRC_CUST_ADD
(W_CN IN NUMBER, W_CLN IN VARCHAR, W_CFN IN VARCHAR, W_CBAL IN NUMBER) AS
BEGIN
INSERT INTO CUSTOMER (CUST_NUM, CUST_LNAME, CUST_FNAME, CUST_BALANCE)
VALUES (W_CN, W_CLN, W_CFN, W_CBAL);
END;
To test the procedure:
EXEC PRC_CUST_ADD(1002,’Rauthor’,’Peter’,0.00);
SELECT * FROM CUSTOMER;
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16. Write a procedure to add a new invoice record to the INVOICE table. Use the following values
in the new record:
8006, 1000, ’30-APR-08’, 301.72
Name the procedure prc_invoice_add. Run a query to see if the record has been
added.
CREATE OR REPLACE PROCEDURE PRC_INVOICE_ADD
(W_IN IN NUMBER, W_CN IN NUMBER, W_ID IN DATE, W_IA IN NUMBER) AS
BEGIN
INSERT INTO INVOICE
VALUES (W_IN, W_CN, W_ID, W_IA);
END;
To test the procedure:
SELECT * FROM CUSTOMER;
SELECT * FROM INVOICE;
EXEC PRC_INVOICE_ADD(8006,1000,’30-APR-08’,301.72);
SELECT * FROM INVOICE;
SELECT * FROM CUSTOMER;
17. Write a trigger to update the customer balance when an invoice is deleted. Name the trigger
trg_updatecustbalance2.
CREATE OR REPLACE TRIGGER TRG_UPDATECUSTBALANCE2
AFTER DELETE ON INVOICE
FOR EACH ROW
BEGIN
UPDATE
CUSTOMER
SET CUST_BALANCE = CUST_BALANCE - :OLD.INV_AMOUNT
WHERE
CUST_NUM = :OLD.CUST_NUM;
END;
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18. Write a procedure to delete an invoice given the invoice number as a parameter. Name the
procedure prc_inv_delete. Test the procedure by deleting invoices 8005 and 8006.
CREATE OR REPLACE PROCEDURE PRC_INV_DELETE (W_IN IN NUMBER) AS
BEGIN
IF W_IN IS NOT NULL THEN
DELETE FROM INVOICE WHERE INV_NUM = W_IN;
END IF;
END;
To test the procedure:
SELECT * FROM CUSTOMER;
SELECT * FROM INVOICE;
EXEC PRC_INV_DELETE(8005);
EXEC PRC_INV_DELETE(8006);
SELECT * FROM INVOICE;
SELECT * FROM CUSTOMER;
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NOTE
The following problem sets can serve as the basis for a class project or case.
Figure P8.19 Ch08_SaleCo2 Database Tables
Use the Ch08_SaleCo2 database to work Problems 19-22.
ONLINE CONTENT
The Ch08_SaleCo2 database used in Problems 19-22 is located in the Online Student
Companion. This database is stored in Microsoft Access format. The Student Online
Companion also includes SQL script files (Oracle and SQLServer) for all of the data sets used
throughout the book.
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19. Create a trigger named trg_line_total to write the LINE_TOTAL value in the LINE table
every time you add a new LINE row. (The LINE_TOTAL value is the product of the
LINE_UNITS and the LINE_PRICE values.)
CREATE OR REPLACE TRIGGER TRG_LINE_TOTAL
BEFORE INSERT ON LINE
FOR EACH ROW
BEGIN
:NEW.LINE_TOTAL := :NEW.LINE_UNITS * :NEW.LINE_PRICE;
END;
20. Create a trigger named trg_line_prod that will automatically update the product quantity on
hand for each product sold after a new LINE row is added.
CREATE OR REPLACE TRIGGER TRG_LINE_PROD
AFTER INSERT ON LINE
FOR EACH ROW
BEGIN
UPDATE PRODUCT
SET
P_QOH = P_QOH - :NEW.LINE_UNITS
WHERE PRODUCT.P_CODE = :NEW.P_CODE;
END;
21. Create a stored procedure named prc_inv_amounts to update the INV_SUBTOTAL,
INV_TAX, and INV_TOTAL. The procedure takes the invoice number as a parameter. The
INV_SUBTOTAL is the sum of the LINE_TOTAL amounts for the invoice, the INV_TAX is
the product of the INV_SUBTOTAL and the tax rate (8%), and the INV_TOTAL is the sum of
the INV_SUBTOTAL and the INV_TAX.
CREATE OR REPLACE PROCEDURE PRC_INV_AMOUNTS (W_IN IN NUMBER) AS
W_CHK NUMBER := 0;
W_SUBT NUMBER := 0;
W_TAX NUMBER := 0;
BEGIN
-- VALIDATE INVOICE NUMBER
SELECT COUNT(*) INTO W_CHK FROM INVOICE
WHERE INV_NUMBER = W_IN;
IF W_CHK = 1 THEN
-- COMPUTE THE W_SUBT
SELECT SUM(LINE_TOTAL) INTO W_SUBT FROM LINE
WHERE LINE.INV_NUMBER = W_IN;
-- COMPUTE W_TAX
W_TAX := W_SUBT * 0.08;
-- UPDATE INVOICE
UPDATE INVOICE
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SET INV_SUBTOTAL = W_SUBT,
INV_TAX = W_TAX,
INV_TOTAL = W_SUBT + W_TAX
WHERE INV_NUMBER = W_IN;
END IF;
END;
22. Create a procedure named prc_cus_balance_update that will take the invoice number as a
parameter and update the customer balance. (Hint: You can use the DECLARE section to
define a TOTINV numeric variable that holds the computed invoice total.)
NOTE
Actually, the TOTINV is not really needed – because the INVOICE table already contains
the INV_TOTAL attribute. The procedure we have shown next uses the INV_TOTAL
attribute.
CREATE OR REPLACE PROCEDURE PRC_CUS_BALANCE_UPDATE (W_IN IN NUMBER) AS
W_CUS NUMBER := 0;
W_TOT NUMBER := 0;
BEGIN
-- GET THE CUS_CODE
SELECT CUS_CODE INTO W_CUS
FROM INVOICE
WHERE INVOICE.INV_NUMBER = W_IN;
-- UPDATES CUSTOMER IF W_CUS > 0
IF W_CUS > 0 THEN
UPDATE CUSTOMER
SET CUS_BALANCE = CUS_BALANCE +
(SELECT INV_TOTAL FROM INVOICE WHERE INV_NUMBER = W_IN)
WHERE CUS_CODE = W_CUS;
END IF;
END;
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Use the Ch08_AviaCo database to work Problems 23-34.
ONLINE CONTENT
The Ch08_AviaCo database used for Problems 23−34 is located in the Online Student
Companion. This database is stored in Microsoft Access format. The Student Online
Companion also includes SQL script files (Oracle and SQLServer) for all of the data sets used
throughout the book.
Figure P8.23 Ch08_AviaCo Database Tables
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Chapter 8 Advanced SQL
23. Modify the MODEL table to add the following attribute and insert the values shown in Table
P8.23.
Table P8.23 The New Attribute for the MODEL Table
Attribute name
MOD_WAIT_CHG
Attribute Description Attribute type
Waiting charge per Numeric
hour for each model:
Attribute Values
$100 for C-90A
$50 for PA23-250
$75 for PA31-350
ALTER TABLE MODEL ADD MOD_WAIT_CHG NUMBER;
UPDATE MODEL
SET MOD_WAIT_CHG = 100
WHERE
MOD_CODE = ‘C-90A’;
UPDATE MODEL
SET MOD_WAIT_CHG = 50
WHERE
MOD_CODE = ‘PA23-250’;
UPDATE MODEL
SET MOD_WAIT_CHG = 75
WHERE
MOD_CODE = ‘PA31-350’;
24. Write the queries to update the MOD_WAIT_CHG attribute values based on problem 23.
UPDATE MODEL SET MOD_WAIT_CHG = 100
UPDATE MODEL SET MOD_WAIT_CHG = 50
UPDATE MODEL SET MOD_WAIT_CHG = 75
WHERE MOD_CODE = 'C-90A';
WHERE MOD_CODE = 'PA23-250';
WHERE MOD_CODE = 'PA31-350';
25. Modify the CHARTER table to add the attributes shown in Table P8.25.
Table P8.25 The New Attributes for the CHARTER Table
Attribute name
CHAR_WAIT_CHG
Attribute Description
Attribute type
Waiting charge for each model (copied from the Numeric
MODEL table.)
CHAR_FLT_CHG_HR Flight charge per mile for each model (copied Numeric
from
the
MODEL
table
using
the
MOD_CHG_MILE attribute.)
CHAR_FLT_CHG
Flight charge (calculated by
Numeric
CHAR_HOURS_FLOWN x CHAR_FLT_CHG_HR)
CHAR_TAX_CHG
CHAR_TOT_CHG
CHAR_PYMT
CHAR_BALANCE
CHAR_FLT_CHG x tax rate (8%)
CHAR_FLT_CHG + CHAR_TAX_CHG
Amount paid by customer
Balance remaining after payment
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Numeric
Numeric
Numeric
Numeric
Chapter 8 Advanced SQL
ALTER TABLE CHARTER
ADD CHAR_WAIT_CHG NUMBER
ADD CHAR_FLT_CHG_HR NUMBER
ADD CHAR_FLT_CHG NUMBER
ADD CHAR_TAX_CHG NUMBER
ADD CHAR_TOT_CHG NUMBER
ADD CHAR_PYMT NUMBER
ADD CHAR_BALANCE NUMBER;
26. Write the sequence of commands required to update the CHAR_WAIT_CHG attribute values
in the CHARTER table. Hint: Use either an updatable view or a stored procedure.
NOTE
This query can be done with an UPDATE statement and a SELECT subquery.
UPDATE CHARTER
SET CHAR_WAIT_CHG = (
SELECT
MOD_WAIT_CHG FROM MODEL, AIRCRAFT
WHERE
MODEL.MOD_CODE = AIRCRAFT.MOD_CODE
AND
AIRCRAFT.AC_NUMBER = CHARTER.AC_NUMBER);
27. Write the sequence of commands required to update the CHAR_FLT_CHG_HR attribute
values in the CHARTER table. Hint: Use either an updatable view or a stored procedure.
NOTE
This query can be done with an UPDATE statement and a SELECT subquery.
UPDATE CHARTER
SET CHAR_FLT_CHG_HR = (
SELECT
MOD_CHG_MILE FROM MODEL, AIRCRAFT
WHERE
MODEL.MOD_CODE = AIRCRAFT.MOD_CODE
AND
AIRCRAFT.AC_NUMBER = CHARTER.AC_NUMBER);
28. Write the command required to update the CHAR_FLT_CHG attribute values in the
CHARTER table.
UPDATE CHARTER
SET CHAR_FLT_CHG = CHAR_HOURS_FLOWN * CHAR_FLT_CHG_HR;
29. Write the command required to update the CHAR_TAX_CHG attribute values in the
CHARTER table.
UPDATE CHARTER
SET CHAR_TAX_CHG = CHAR_FLT_CHG * 0.08;
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30. Write the command required to update the CHAR_TOT_CHG attribute values in the
CHARTER table.
UPDATE CHARTER
SET CHAR_TOT_CHG = CHAR_FLT_CHG + CHAR_TAX_CHG;
31. Modify the PILOT table to add the attribute shown in Table P8.21.
Table P8.31 The New Attribute for the PILOT Table
Attribute name
PIL_PIC_HRS
Attribute Description
Attribute type
Pilot in command (PIC) hours. Updated by adding the Numeric
CHARTER table’s CHAR_HOURS_FLOWN to the
PIL_PIC_HRS when the CREW table shows the
CREW_JOB to be pilot.
ALTER TABLE PILOT ADD PIL_PIC_HRS NUMBER;
32. Create a trigger named trg_char_hours that will automatically update the AIRCRAFT table
when a new CHARTER row is added. Use the CHARTER table’s CHAR_HOURS_FLOWN
to update the AIRCRAFT table’s AC_TTAF, AC_TTEL, and AC_TTER values.
CREATE OR REPLACE TRIGGER TRG_CHAR_HOURS
AFTER INSERT ON CHARTER
FOR EACH ROW
BEGIN
UPDATE AIRCRAFT
SET AC_TTAF = AC_TTAF + :NEW.CHAR_HOURS_FLOWN,
AC_TTEL = AC_TTEL + :NEW.CHAR_HOURS_FLOWN,
AC_TTER = AC_TTER + :NEW.CHAR_HOURS_FLOWN
WHERE AIRCRAFT.AC_NUMBER = :NEW.AC_NUMBER;
END;
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33. Create a trigger named trg_pic_hours that will automatically update the PILOT table when a
new CREW row is added and the CREW table uses a ‘pilot’ CREW_JOB entry. Use the
CHARTER table’s CHAR_HOURS_FLOWN to update the PILOT table’s PIL_PIC_HRS
only when the CREW table uses a ‘pilot’ CREW_JOB entry.
CREATE OR REPLACE TRIGGER TRG_PIC_HOURS
AFTER INSERT ON CREW
FOR EACH ROW
BEGIN
IF :NEW.CREW_JOB = 'Pilot' THEN
UPDATE PILOT
SET PIL_PIC_HRS = PIL_PIC_HRS +
(SELECT CHAR_HOURS_FLOWN FROM CHARTER
WHERE CHAR_TRIP = :NEW.CHAR_TRIP)
WHERE EMP_NUM = :NEW.EMP_NUM;
END IF;
END;
34. Create a trigger named trg_cust_balance that will automatically update the CUSTOMER
table’s CUST_BALANCE when a new CHARTER row is added. Use the CHARTER table’s
CHAR_TOT_CHG as the update source (Assume that all charter charges are charged to the
customer balance.)
CREATE OR REPLACE TRIGGER TRG_CUST_BALANCE
AFTER INSERT ON CHARTER
FOR EACH ROW
BEGIN
UPDATE CUSTOMER
SET CUS_BALANCE = CUS_BALANCE + :NEW.CHAR_TOT_CHG
WHERE CUSTOMER.CUS_CODE = :NEW.CUS_CODE;
END;
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