Chapter Six Introduction to Structured Query Language (SQL)
ANSWERS TO GROUP I QUESTIONS
6.1
What does SQL stand for?
Structured Query Language
6.2
What is a data sublanguage?
A data sublanguage consists only of language statements for defining and processing a database. To
obtain a full programming language, SQL statements must be embedded in scripting languages such as
VBScript or in programming languages such as Java or C#.
6.3
Explain the importance of SQL-92.
This is the version of SQL supported by most commonly used database management systems. The
SQL-92 version includes some object-oriented concepts but is not commonly implemented.
6.4
Why is it important to learn SQL?
Most, but not all, DBMS provide graphical tools to provide access to the database. You cannot do
everything with graphical tools that you can do with SQL; furthermore, you must use SQL to generate
SQL statements dynamically in program code.
6.5
In your own words, describe the meanings of the phrases ON UPDATE CASCADE and ON
DELETE CASCADE.
Both are used to support referential integrity actions. On Update Cascade says that if the primary key
of a parent is changed, also change the foreign key in all of that parent’s children. On Delete Cascade
says if a parent is deleted, also delete all of that parent’s children.
Use the following tables for your answers to questions 6.6-6.9:
EMPLOYEE (EmployeeID, Name, Phone, Email)
COMPUTER (SerialNumber, Make, Model, EmployeeID)
(See Figure 6-8 for sample data.)
6.6
Code a CREATE TABLE command to create the EMPLOYEE table. Justify your choices of
column data types and constraints.
Create Table EMPLOYEE
(EmployeeID
Integer,
Name
Char(20)
Not Null,
Phone
Char(12)
Unique Not Null,
Email
Char(20)
Unique Not Null,
CONSTRAINT EmployeePK PRIMARY KEY (EmployeeID));
EmployeeID is defined as an Integer because all Employee IDs are numbers and have a value of 400
or less. EmployeeID is also defined as the primary key as shown in the table description. All other
attributes (including the Phone) contain characters so they are defined as character attributes. All
attributes are Not Null because, based upon the data shown, each employee has a value for the
attribute. Phone and Email are defined as unique because they should be alternate keys for an
employee.
6-1
Chapter Six Introduction to Structured Query Language (SQL)
6.7
Code a CREATE TABLE command to create the COMPUTER table. Justify your choices of
column constraints. Should EmployeeID be null or not null?
Create Table COMPUTER
(SerialNumber
Integer,
Make
Char(20)
Not Null,
Model
Integer
Not Null,
EmployeeID
Integer
Null,
CONSTRAINT ComputerPK PRIMARY KEY (SerialNumber),
CONSTRAINT EmployeeFK FOREIGN KEY (EmployeeID) References
EMPLOYEE(EmployeeID));
SerialNumber is defined as an integer because all serial numbers are numbers and have a value of
8000 or less. SerialNumber is also defined as the primary key as shown in the table description. Make
contains characters so it is defined as a character attribute. Model contains numbers so it is an integer.
EmployeeId is defined as a foreign key to Employee. It is an integer to make it domain compatible
with the Employee table’s primary key. It is allowed to be Null because some Computers may not be
assigned to employees.
6.8
Create a referential integrity constraint on COMPUTER.EmployeeID. Assume that deletions
should not cascade, but that updates should cascade.
Change the EmployeeFK constraint to:
CONSTRAINT EmployeeFK FOREIGN KEY (EmployeeID) References EMPLOYEE(EmployeeID)
ON UPDATE CASCADE);
6.9
Code the required SQL statements for the following alternative version of the COMPUTER
table:
COMPUTER1 (SerialNumber, Make, Model, EmployeeID)
Create Table COMPUTER1
(SerialNumber
Integer,
Make
Char(20)
Not Null,
Model
Integer
Not Null,
EmployeeID
Integer,
CONSTRAINT ComputerPK PRIMARY KEY (SerialNumber, EmployeeID)
CONSTRAINT EmployeeFK FOREIGN KEY (EmployeeID) References
EMPLOYEE(EmployeeID));
6.10 Is COMPUTER or COMPUTER1 a better design? Explain your rationale.
Depends upon the requirements. If there are computers that are not assigned to an Employee, the
COMPUTER definition is better. If all computers must be assigned to an employee before their
information is stored in the database, COMPUTER1 is acceptable, but not positive that it is better.
6.11 Code SQL to remove the EMPLOYEE table from the database. Assume that the referential
integrity constraint is to be removed.
ALTER TABLE ASSIGNMENT DROP CONSTRAINT ComputerFK;
DROP TABLE EMPLOYEE.
6-2
Chapter Six Introduction to Structured Query Language (SQL)
6.12 Code SQL to remove the EMPLOYEE table from the database. Assume that the
COMPUTER table is to be removed, also.
DROP TABLE COMPUTER.
DROP TABLE EMPLOYEE.
For questions 6.13–6.42, use the EMPLOYEE and COMPUTER tables. Assume that EmployeeID
and Computer ID are surrogate keys.
6.13 Write a SQL statement to display all columns of all rows of COMPUTER. Do not use the *
notation.
SELECT SerialNumber, Make, Model, EmployeeID
FROM COMPUTER;
6.14 Write a SQL statement to display all columns of all rows of COMPUTER. Use the * notation.
SELECT *
FROM COMPUTER;
6.15 Write SQL to display the Make and Model of all COMPUTERs.
SELECT Make, Model
FROM COMPUTER;
6.16 Write SQL to display the SerialNumber, Make, and Model of all COMPUTERs having the
Make ‘Dell’.
SELECT SerialNumber, Make, Model
FROM COMPUTER
WHERE Make = ‘Dell’;
6.17 Write SQL to display the Make column of COMPUTER.
SELECT Make
FROM COMPUTER;
6.18 Write SQL to display the Make column of COMPUTER (do not show duplicates).
SELECT Distinct Make
FROM COMPUTER;
6.19 Write SQL to display the Make and Model of all computers having the type ‘Dell’ and the
Model is ’Laptop.’
SELECT Make, Model
FROM COMPUTER
WHERE Make = ‘Dell’ AND
Model = 4100;
6.20 Write SQL to display the ComputerID, Make, and Model of all computers that are not of Type
Dell or IBM.
SELECT SerialNumber, Make, Model
FROM COMPUTER
WHERE Make = ‘Dell’;
6-3
Chapter Six Introduction to Structured Query Language (SQL)
6.21 Write SQL to display the ComputerID, Make, and Model for all computers having a fourcharacter Type starting with D.
SELECT SerialNumber, Make, Model
FROM COMPUTER
WHERE Make LIKE ‘D_ _ _’;
6.22 Write SQL to display the Name and Email address of all employees who have an email
address ending with ‘somewhere.com’. Assume that email account names can be any
number of characters.
SELECT Name, Email
FROM EMPLOYEE
WHERE Email LIKE ‘%@somewhere.com’;
6.23 Write SQL to display the ComputerID of any computer that has a null value for Model.
SELECT SerialNumber
FROM COMPUTER
WHERE Model IS NULL’;
6.24 Write SQL to display the Make and Model of all computers sorted by Name.
SELECT Make, Model
FROM COMPUTER
ORDER BY Make;
6.25 Write SQL to display the Make and Model of all computers sorted by Make in ascending
order and by Model in descending order within Make.
SELECT Make, Model
FROM COMPUTER
ORDER BY Make ASC Model DESC;
6.26 Write SQL to count the number of computers.
SELECT COUNT(*)
FROM COMPUTER;
6.27 Write SQL to count the number of distinct Makes.
SELECT COUNT(DISTINCT MAKE)
FROM COMPUTER;
6.28 For the following table, write SQL to display the minimum, maximum, and average Weight of
computers.
COMPUTER2 (SerialNumber, Make, Model, Weight, OwnerID)
SELECT MIN(Weight), MAX(Weight), AVG(Weight)
FROM COMPUTER;
6.29 For the COMPUTER2 table, write SQL to group the data by Make and display the average
Weight per make.
SELECT Make, AVG(Weight)
6-4
Chapter Six Introduction to Structured Query Language (SQL)
FROM COMPUTER
GROUP BY Make;
6.30 Answer question 6.29, but consider only Makes for which there are five or more computers in
the database.
SELECT Make, AVG(Weight)
FROM COMPUTER
GROUP BY Make
HAVING Count(Make) > 5;
6.31 Answer question 6.30, but do not consider any computer having the Make ‘Dell’.
SELECT Make, AVG(Weight)
FROM COMPUTER
WHERE NOT Make = ‘Dell’
GROUP BY Make
HAVING Count(Make) > 5;
6.32 Using the tables in Figure 6-8, write SQL to display Name and Email of any employees with
Dell computers. Use a subquery.
SELECT Name, Email
FROM EMPLOYEE
WHERE EmployeeID IN
(SELECT EmployeeID
FROM COMPUTER
WHERE Make = ‘Dell’);
6.33 Using the tables in Figure 6-8, write SQL to display Name and Email of any employees with
IBM computers with the Model ‘Laptop’. Use a join.
SELECT Name, Email
FROM EMPLOYEE, COMPUTER
WHERE EMPLOYEE.EmployeeID = COMPUTER.EmployeeID
AND Make = ‘IBM’
AND Model = ‘Laptop’;
6.34 Suppose that the following new table is added to the database:
MANUFACTURER (Name, City, State)
Assume that Make of COMPUTER is a foreign key into Name of MANUFACTURER. Code
SQL to display the Name and Email of any employee with a computer that was made by a
manufacturer in Atlanta. Use subquery format.
SELECT Name, Email
FROM EMPLOYEE
WHERE EmployeeID IN
(SELECT EmployeeID
FROM COMPUTER
WHERE Make IN
(SELECT Name
FROM MANUFACTURER
6-5
Chapter Six Introduction to Structured Query Language (SQL)
WHERE City = ‘Atlanta’)));
6.35 Answer question 6.34, but use joins in the standard SELECT FROM WHERE format.
SELECT Name, Email
FROM EMPLOYEE, COMPUTER, MANUFACTURER
WHERE EMPLOYEE.EmployeeID = COMPUTER.EmployeeID
AND COMPUTER.Make = MANUFACTURER.Name;
6.36 Answer question 6.34, but use joins in the JOIN ON format.
SELECT Name, Email
FROM ((EMPLOYEE JOIN COMPUTER
ON EMPLOYEE.EmployeeID = COMPUTER.EmployeeID)
JOIN MANUFACTURER
ON COMPUTER.Make = MANUFACTURER.Name);
6.37 Answer question 6.34, but include employees, regardless of whether they have a computer.
SELECT Name, Email
FROM ((EMPLOYEE LEFT JOIN COMPUTER
ON EMPLOYEE.EmployeeID = COMPUTER.EmployeeID)
LEFT JOIN MANUFACTURER
ON COMPUTER.Make = MANUFACTURER.Name);
6.38 Code SQL to add three new rows to the COMPUTER table. Assume that ComputerID is a
surrogate key and the DBMS will provide a value for it. Otherwise, assume you have all of the
data.
INSERT INTO COMPUTER VALUES (‘IBM’, 2100, 100);
INSERT INTO COMPUTER VALUES (‘IBM’, 330, 100);
INSERT INTO COMPUTER VALUES (‘Gateway’, 6690, 300);
6.39 Code SQL to add three new rows to the EMPLOYEE table. Assume that EmployeeID is a
surrogate key and the DBMS will provide a value for it. Otherwise, you have only values for
Name and Phone; and assume that Email is Not Null.
INSERT INTO EMPLOYEE (Name, Phone, Email) VALUES
(‘Jack Jones’, ‘540-433-0099’, NULL);
INSERT INTO EMPLOYEE (Name, Phone, Email) VALUES
(‘Jill Jones’, ‘540-433-2233’, NULL);
INSERT INTO EMPLOYEE (Name, Phone, Email) VALUES
(‘Jacoline Jenkins, ‘777-888-9999’, NULL);
6-6
Chapter Six Introduction to Structured Query Language (SQL)
6.40 Code SQL to change the values of ‘Compaq’ in Make of COMPUTER to ‘HP’.
UPDATE COMPUTER
SET Make = ‘HP’
WHERE Make = ‘Compaq’;
6.41 Explain what will happen if you leave the WHERE clause off your answer to question 6.40.
The Make of ALL computers would be changed to ‘HP’.
6.42 Code SQL to delete all rows of COMPUTERs of Model ‘Desktop’. What will happen if you
forget to code the WHERE clause in this statement?
DELETE FROM COMPUTER
WHERE Model = ‘Desktop’;
If you leave off the where clause, all rows will be deleted. The structure of the table will remain, but
the table will be empty.
 ANSWERS TO GROUP II QUESTIONS
6.43 Write SQL statements to create the tables for a relational design of the data model shown in
Figure 5-17(c). Use the format of statements shown in Figure 6-7. Define foreign keys and
set cascading behavior in accordance with the referential integrity actions shown there.
Which of the referential integrity actions cannot be implemented using foreign key
constraints?
CREATE TABLE SALESPERSON
(EmployeeID
Integer
Not Null,
Name
Char(20)
Not Null,,
Phone
Char(12)
Not Null,,
Email
Char(20)
Not Null,,
CONSTRAINT EmployeePK PRIMARY KEY (EmployeeID));
CREATE TABLE SALESORDER
(SalesOrderNumber Integer
Not Null,
OrderDate
DATE
Not Null,
OrderDescription
VarChar(50)
Not Null,,
OrderTotal
Number(8,2)
Not Null,,
TotalCommission
Number(6,2)
Not Null,,
CONSTRAINT EmployeePK PRIMARY KEY (EmployeeID));
CREATE TABLE SALESPERSON_SHARE
(EmployeeID
Integer
Not Null,
SalesOrderNumber Integer
Not Null,
CommissionPercent Number(3,2)
Not Null,,
CONSTRAINT SharePK PRIMARY KEY (EmployeeID, SalesOrderNumber));
ALTER TABLE SALESPERSON_SHARE
ADD CONSTRAINT EmployeeFK
FOREIGN KEY (EmployeeID) REFERENCES EMPLOYEE (EmployeeID)
ON UPDATE CASCADE
6-7
Chapter Six Introduction to Structured Query Language (SQL)
ON DELETE NO ACTION;
ALTER TABLE SALESPERSON_SHARE
ADD CONSTRAINT BookFK
FOREIGN KEY (SalesOrderNumber) REFERENCES SALESORDER (SalesOrderNumber)
ON UPDATE CASCADE
ON DELETE CASCADE;
6.44 Use SQL INSERT statements to add data to the tables you created in Question 6.43. Ensure
that you have at least three salespeople, three sales orders, and seven records of
salesperson share.
Insert Into SALESPERSON Values (10, ‘Jack Jones’, ‘540.111.2222’, ‘JackJ@problems.com’);
Insert Into SALESPERSON Values (20, ‘Jill Jones’, ‘540.333.4444’, ‘JillJ@problems.com’);
Insert Into SALESPERSON Values (30, ‘Bill Baker’, ‘540.555.6666’, ‘BillB@problems.com’);
Insert Into SALESORDER Values
(101, ’5-10-2003’, ‘Discount Order for PeakView’, 2500.00, 250.00);
Insert Into SALESORDER Values
(102, ’5-22-2003’, ‘Replace Playground Equipment’, 55000.00, 5500.00);
Insert Into SALESORDER Values
(103, ’6-1-2003’, ‘Refurnish Hotel Lobby’, 100000, 10000.00);
Insert Into SALESPERSON_SHARE Values (10, 101, 1.00);
Insert Into SALESPERSON_SHARE Values (10, 102, .25);
Insert Into SALESPERSON_SHARE Values (20, 102, .25);
Insert Into SALESPERSON_SHARE Values (30, 102, .50);
Insert Into SALESPERSON_SHARE Values (20, 103, .60);
Insert Into SALESPERSON_SHARE Values (30, 103, .40);
6.45 Using the tables from your answers to questions 6.43 and 6.44, write SQL statements for the
following questions. Where joins are required, use the JOIN ON format of join; use aliases for
table names in join statements.
A. Show the Names and Phone numbers of SALESPEOPLE who have a null Email address.
Explain why the query should find no such rows.
SELECT Name, Phone
FROM SALESPERSON
WHERE Email Is Null;
No rows should be found because all Salespeople have an Email Address.
B. Show the Names of SALESPEOPLE and the total and average of OrderTotal of all orders
in which they have participated.
SELECT SP.Name, Sum(SO.OrderTotal), Avg(SO.OrderTotal)
FROM SALESPERSON AS SP INNER JOIN (SalesOrder AS SO INNER JOIN
SalesPerson_Share AS SS ON SO.SalesOrderNumber = SS.SalesOrderNumber) ON
SP.EmployeeID = SS.EmployeeID
GROUP BY SP.Name;
6-8
Chapter Six Introduction to Structured Query Language (SQL)
C. For each order, show the SalesOrderNumber and sum of CommissionPercent.
SELECT SalesOrderNumber, SUM(CommissionPercent)
FROM SALESPERSON_SHARE
GROUP BY SalesOrderShare;
D. Show the SalesOrderNumber and Name of SALESPERSON for orders in which more
than two salespeople have participated.
SELECT SP.Name,SS.SalesOrderNumber
FROM SALESPERSON AS SP INNER JOIN (SALESORDER AS SO
INNER JOIN salesperson_share AS SS ON SO.SalesOrderNumber =
SS.SalesOrderNumber) ON SP.EmployeeID = SS.EmployeeID
WHERE SS.SalesOrderNumber IN
(SELECT SALESPERSON_SHARE.SalesOrderNumber
FROM SALESPERSON_SHARE
GROUP BY SALESPERSON_SHARE.SalesOrderNumber
HAVING ((Count(SALESPERSON_SHARE.EmployeeID)>2)));
E. Same as B, but include all salespeople, regardless of whether they have any
SALESPERSON_SHARE data.
SELECT Name, SUM (OrderTotal), AVG (OrderTotal)
FROM ((SALESPEOPLE AS SP LEFT JOIN SALESPERSON_SHARE AS SS
ON SP.EmployeeId = SS.EmployeeID)
RIGHT JOIN SALESORDER AS SO ON SO.SalesOrderNumber =
SS.SalesOrderNumber)
GROUP BY Name;
F. It is possible to perform calculations in SELECT statements. Thus, the following displays
the product of columns A and B as a new column C:
SELECT A, B, (A*B) as C
FROM TABLE1;
Using this structure, code a SQL statement to list the Names of SalesPeople and the
amount of commission they have received on each order in which they have participated.
SELECT Name, SS.CommissionPercent * SO.OrderTotal
FROM ((SALESPERSON AS SP LEFT JOIN SALESPERSON_SHARE AS SS
ON SP.EmployeeId = SS.EmployeeID)
LEFT JOIN SALESORDER AS SO ON SO.SalesOrderNumber = SS.SalesOrderNumber);
6.46 Write SQL statements to create the tables shown in Figure 5-20(b). Drop the word THREE
from the first two tables; it adds no value here. Thus, the first table should be named
HOTEL_INVOICE. Use the format of statements shown in Figure 6-7. Define foreign keys
and set cascading behavior in accordance with the referential integrity actions shown there.
Are there referential integrity actions that cannot be implemented using foreign key
constraints?
6-9
Chapter Six Introduction to Structured Query Language (SQL)
CREATE TABLE HOTEL_INVOICE
(InvoiceNumber
Integer
Not Null,
Arrival_Date
Date
Not Null,
CustomerName
Char(20)
Not Null,
Total
Numeric (6,2),
CONSTRAINT InvoiceNumberPK PRIMARY KEY (InvoiceNumber);
CREATE TABLE INVOICE_LINE_ITEM
(ItemDate
Date
Not Null,
ItemCategory
Char(20)
Not Null,
InvoiceNumber
Integer
Not Null,
CategoryTotal
Numeric (6,2),
CONSTRAINT InvoiceLinePK PRIMARY KEY (ItemDate, ItemCategory, InvoiceNumber);
CREATE TABLE SUB_LINE_ITEM
(SubCategoryName
Char(10)
Not Null,
ItemDate
Date
Not Null,
ItemCategory
Char(20)
Not Null,
InvoiceNumber
Integer
Not Null,
SubCategoryCost
Numeric (6,2),
CONSTRAINT SubLinePK PRIMARY KEY (SubCategoryName, ItemDate, ItemCategory,
InvoiceNumber);
ALTER TABLE INVOICE_LINE_ITEM
ADD CONSTRAINT InvoiceFK
FOREIGN KEY (InvoiceNumber) REFERENCES HOTEL_INVOICE (InvoiceNumber)
ON UPDATE CASCADE
ON DELETE CASCADE;
ALTER TABLE SUB_LINE_ITEM
ADD CONSTRAINT Line_Item_FK
FOREIGN KEY (ItemDate, ItemCategory, InvoiceNumber) REFERENCES
INVOICE_LINE_ITEM (ItemDate, ItemCategory, InvoiceNumber)
ON UPDATE CASCADE
ON DELETE CASCADE;
All referential integrity constraints shown can be implemented using foreign keys.
6.47 Use SQL INSERT statements to add data to the tables you created in Question 6.46. Ensure
that you have at least three HOTEL_INVOICE rows, six INVOICE_LINE_ITEM rows, and
twelve SUB_LINE_ITEM rows. Two invoices should be for the same customer. Also, at least
one of your invoices should have no INVOICE_LINE_ITEM rows, and at least one of your
INVOICE_LINE_ITEM rows should have no SUB_LINE_ITEM rows.
Insert Into HOTEL_INVOICE Values (10, ‘5-10-2003’, ‘Jill Jones’, 500.00);
Insert Into HOTEL_INVOICE Values (20, ‘5-10-2003’, ‘Jack Jones’, NULL);
Insert Into HOTEL_INVOICE Values (30, ‘5-12-2003’, ‘Jenny Jones’, 100.00);
Insert Into INVOICE_LINE_ITEM Values (‘5-10-2003’, ‘Restaurant’, 10, 25.95);
Insert Into INVOICE_LINE_ITEM Values (‘5-10-2003’, ‘Room’, 10, 120.00);
Insert Into INVOICE_LINE_ITEM Values (‘5-10-2003’, ‘Tax’, 10, 22.50);
Insert Into INVOICE_LINE_ITEM Values (‘5-11-2003’, ‘Restaurant’, 10, 55.00);
6-10
Chapter Six Introduction to Structured Query Language (SQL)
Insert Into INVOICE_LINE_ITEM Values (‘5-12-2003’, ‘Room’, 30, 120.00);
Insert Into INVOICE_LINE_ITEM Values (‘5-12-2003’, ‘Tax’, 30, 22.50);
Insert Into SUB_LINE_ITEM Values (‘Breakfast’, ‘5-10-2003’, ‘Restaurant’, 10, 5.95);
Insert Into SUB_LINE_ITEM Values (‘Lunch’, ‘5-10-2003’, ‘Restaurant’, 10, 7.00);
Insert Into SUB_LINE_ITEM Values (‘Dinner’, ‘5-10-2003’, ‘Restaurant’, 10, 13.00);
Insert Into SUB_LINE_ITEM Values (‘State’, ‘5-10-2003’, ‘Tax’, 10, 10.00);
Insert Into SUB_LINE_ITEM Values (‘Local’, ‘5-10-2003’, ‘Tax’, 10, 12.50);
Insert Into SUB_LINE_ITEM Values (‘Breakfast’, ‘5-11-2003’, ‘Restaurant’, 10, 10.00);
Insert Into SUB_LINE_ITEM Values (‘Lunch’, ‘5-10-2003’, ‘Restaurant’, 10, 7.00);
Insert Into SUB_LINE_ITEM Values (‘Room Service’, ‘5-10-2003’, ‘Restaurant’, 10, 13.00);
Insert Into SUB_LINE_ITEM Values (‘Dinner’, ‘5-11-2003’, ‘Restaurant’, 10, 15.00);
Insert Into SUB_LINE_ITEM Values (‘Breakfast’, ‘5-11-2003’, ‘Restaurant’, 10, 10.00);
Insert Into SUB_LINE_ITEM Values (‘State’, ‘5-12-2003’, ‘Tax’, 30, 10.00);
Insert Into SUB_LINE_ITEM Values (‘Local’, ‘5-12-2003’, ‘Tax’, 30, 12.50);
6.48 Using the tables from your answers to questions 6.46 and 6.47, write SQL statements for the
following questions. Use subqueries except where joins are required. For joins, use the JOIN
ON format and use aliases for table names in join statements.
A. Show the InvoiceNumber, CustomerName for all invoices having a CategoryTotal greater
than 100.
SELECT InvoiceNumber, CustomerName
FROM HOTEL_INVOICE
WHERE InvoiceNumber IN
(SELECT InvoiceNumber
FROM INVOICE_LINE_ITEM
WHERE CategoryTotal > 100);
B. Suppose that you want to show the InvoiceNumber, CustomerName, and sum of
CategoryTotal for each invoice. A subquery will not work. Explain why.
The subquery would need to generate the sum of the CategoryTotal. This sum could not be
displayed if it comes from an inner subquery. You can only display data from the outside query in
a subquery.
C. Construct a join to show the InvoiceNumber, CustomerName, and sum of CategoryTotal
for each invoice.
SELECT InvoiceNumber, CustomerName, SUM(CategoryTotal)
FROM HOTEL_INVOICE JOIN INVOICE_LINE_ITEM ON
HOTEL_INVOICE.InvoiceNumber = INVOICE_LINE_ITEM.InvoiceNumber
GROUP BY HOTEL_INVOICE.InvoiceNumber;
D. Construct a join to show the CustomerName, and sum of CategoryTotal for each
customer.
SELECT CustomerName, SUM(CategoryTotal)
FROM HOTEL_INVOICE JOIN INVOICE_LINE_ITEM ON
HOTEL_INVOICE.InvoiceNumber = INVOICE_LINE_ITEM.InvoiceNumber
GROUP BY CustomerName;
6-11
Chapter Six Introduction to Structured Query Language (SQL)
E. Construct a join to show the InvoiceNumber, CategoryTotal, and sum of
SubCategoryCost for each line item.
SELECT INVOICE_LINE_ITEM.InvoiceNumber, CategoryTotal, SUM(SubCategoryCost)
FROM INVOICE_LINE_ITEM JOIN SUB_LINE_ITEM ON
(INVOICE_LINE_ITEM.InvoiceNumber = SUB_LINE_ITEM.InvoiceNumber)
AND (INVOICE_LINE_ITEM.ItemDate = SUB_LINE_ITEM.ItemDate)
AND (INVOICE_LINE_ITEM.ItemCategory = SUB_LINE_ITEM.ItemCategory)
GROUP BY INVOICE_LINE_ITEM.InvoiceNumber, CategoryTotal;
F. Construct a join to show the InvoiceNumber, CustomerName, sum of CategoryTotal, and
sum of SubCategoryCost for each invoice.
Since you are trying to sum from two different tables, the join tries to leave duplicate rows for the
Invoice_line_Item, then these duplicates are summed. The Category Total will be added for multiple
times, once for each SubCategoryCost..
 ANSWERS TO FIREDUP PROJECT QUESTIONS
Assume that FiredUp has created a database with the following tables:
CUSTOMER (CustomerSK, Name, Phone, EmailAddress)
STOVE (SerialNumber, Type, Version, DateOfManufacture)
REGISTRATION (CustomerSK, SerialNumber, Date)
STOVE_REPAIR (RepairInvoiceNumber, SerialNumber, Date, Description, Cost, CustomerSK)
Code SQL for the following. Assume that all dates are in the format mmddyyyy.
A. Show all of the data in each of the four FiredUp tables.
SELECT *
FROM CUSTOMER;
SELECT *
FROM STOVE;
SELECT *
FROM REGISTRATION;
SELECT *
FROM STOVE_REPAIR;
B. List the Versions of all stoves.
SELECT Version
FROM STOVE;
To list each very only once use:
SELECT DISTINCT Version
FROM STOVE;
6-12
Chapter Six Introduction to Structured Query Language (SQL)
C. List the Versions of all stoves of the type ‘Fired Now’.
SELECT Version
FROM STOVE
WHERE Type = ‘Fired Now’;
D. List the SerialNumber and Date of all registrations in the year 2002.
SELECT SerialNumber, Date
FROM REGISTRATION
WHERE Date Like ‘%2002’;
E. List the SerialNumber and Date of all registrations in February. Use the underscore (_)
wildcard.
SELECT SerialNumber, Date
FROM REGISTRATION
WHERE Date Like ‘02_ _ _ _ _ _;
F. List the SerialNumber and Date of all registrations in February. Use the percent (%)
wildcard.
SELECT SerialNumber, Date
FROM REGISTRATION
WHERE Date Like ‘02%’;
G. List the names and email addresses of all customers who have an email address.
SELECT Name, EmailAddress
FROM CUSTOMER
WHERE EmailAddress IS NOT NULL;
H. List the names of all customers who do not have an EmailAddress; present the results in
descending sorted order of Name.
SELECT Name, EmailAddress
FROM CUSTOMER
WHERE EmailAddress IS NULL
ORDER BY Name;
I.
Determine the maximum cost of a stove repair.
SELECT MAX (Cost)
FROM STOVE_REPAIR;
J. Determine the average cost of a stove repair.
SELECT AVG (Cost)
FROM STOVE_REPAIR;
K. Count all stoves.
SELECT COUNT (*)
FROM STOVE;
6-13
Chapter Six Introduction to Structured Query Language (SQL)
L. Count all stoves of each type and display the Type and count.
SELECT Type, COUNT (*)
FROM STOVE
GROUP BY Type;
M. List the name and email addresses of all customers who have had a stove repair that
cost more than $50. Use a subquery.
SELECT Name, EmailAddress
FROM CUSTOMER
WHERE CustomerSK IN
(SELECT CustomerSK
FROM STOVE_REPAIR
WHERE Cost > 50.00);
N. List the names and email addresses of all customers who have registered a stove of the
type ‘FiredNow’. Use a subquery.
SELECT Name, EmailAddress
FROM CUSTOMER
WHERE CustomerSK IN
(SELECT CustomerSK
FROM REGISTRATION
WHERE SerialNumber IN
(SELECT SerialNumber
FROM STOVE
WHERE Type = ‘FiredNow’);
O. List the names and email addresses of all customers who have had a stove repair that
cost more than $50. Use join with the JOIN ON syntax.
SELECT Name, EmailAddress
FROM CUSTOMER JOIN (REGISTRATION JOIN STOVE_REPAIR ON
REGISTRATION.SerialNumber = STOVE_REPAIR.Serial_Number) ON
CUSTOMER.CustomerSK = REGISTRATION.CustomerSK
WHERE STOVE_REPAIR.Cost > 50.00;
P. List the names and email addresses of all customers who have registered a stove of type
‘FiredNow’. Use join with the JOIN ON syntax.
SELECT Name, EmailAddress
FROM CUSTOMER JOIN REGISTRATION ON CUSTOMER.CustomerSK =
REGISTRATION.CustomerSK
WHERE STOVE_REPAIR.Type = ‘FiredNow’;
Q. List the names, email addresses, and registration date of all customer registrations.
SELECT Name, EmailAddress, Date
FROM CUSTOMER, REGISTRATION
WHERE CUSTOMER.CustomerSK = REGISTRATION.CustomerSK;
6-14
Chapter Six Introduction to Structured Query Language (SQL)
R. Show the names and email addresses of all customers who have registered a stove, and
have had any stove repaired.
SELECT Name, EmailAddress
FROM CUSTOMER
WHERE CustomerSK IN
(SELECT CustomerSK
FROM REGISTRATION)
AND CustomerSK IN
(SELECT CustomerSK
FROM STOVE_REPAIR);
S. Show the names and email addresses of all customers who have registered a stove, but
who have not had any stove repaired.
SELECT Name, EmailAddress
FROM CUSTOMER
WHERE CustomerSK IN
(SELECT CustomerSK
FROM REGISTRATION)
AND CustomerSK NOT IN
(SELECT CustomerSK
FROM STOVE_REPAIR);
 ANSWERS TO TWIGS TREE TRIMMING SERVICE
Assume that Twigs has created a database with the following tables:
OWNER (OwnerName, Phone, Street, City, State, Zip)
SERVICE (DateOfService, OwnerName, Description, AmountBilled, AmountPaid, DateOfPayment)
CHIP_DELIVERY (CustomerName, DateDelivered, LoadSize, AmountBilled, AmountPaid,
DateOfPayment)
Code SQL for the following. Assume that SERVICE.OwnerName and
CHIP_DELIVERY.CustomerName are foreign keys to OWNER.OwnerName. Also assume that all
dates are in the format mmddyyyy.
A. Show all of the data in each of the three Twigs tables.
SELECT *
FROM OWNER;
SELECT *
FROM SERVICE;
SELECT *
FROM CHIP_DELIVERY;
B. List the Cities of all owners.
SELECT City
FROM OWNER;
6-15
Chapter Six Introduction to Structured Query Language (SQL)
C. List the Cities of all owners without duplicates.
SELECT DISTINCT City
FROM OWNER;
D. List the OwnerName and DateOfService for all services performed in year 2002.
SELECT OwnerName, DateOfService
FROM SERVICE
WHERE LIKE ‘%2002’;
E. List the CustomerName and DateDelivered for all chip deliveries paid for in February.
Use the underscore (_) wildcard.
SELECT OwnerName, DateOfService
FROM SERVICE
WHERE LIKE ’02 _ _ _ _ _ _’;
F. Same as question E, but use the percent (%) wildcard.
SELECT OwnerName, DateOfService
FROM SERVICE
WHERE LIKE ‘02%’;
G. List the names and cities of all customers who have a phone number.
SELECT OwnerName, City
FROM OWNER
WHERE Phone IS NOT NULL;
H. List the names of all customers who do not have a phone number and present the results
in descending sorted order of Name.
SELECT OwnerName, City
FROM OWNER
WHERE Phone IS NULL
ORDER BY OwnerName;
I.
Determine the maximum amount billed for a service.
SELECT MAX (AmountBilled)
FROM SERVICE’;
J. Determine the average load size of a chip delivery.
SELECT AVG (LoadSize)
FROM CHIP_DELIVERY;
K. Count all chip deliveries.
SELECT COUNT (*)
FROM CHIP_DELIVERY;
6-16
Chapter Six Introduction to Structured Query Language (SQL)
L. Count all chip deliveries for each customer and display the CustomerName and count.
SELECT CustomerName, COUNT (*)
FROM CHIP_DELIVERY
GROUP BY CustomerName;
M. List the names and phone numbers of all customers who have service with a non-null
AmountBilled and a null AmountPaid. Use a subquery.
SELECT OwnerName, Phone
FROM OWNER
WHERE OwnerName IN
(SELECT OwnerName
FROM SERVICE
WHERE AmountBilled IS NOT NULL
AND AmountPaid IS NULL);
N. List the names and LoadSize all customers who have a chip delivery and who also have
a service with a non-null AmountBilled and a null AmountPaid. Use a sub-query.
SELECT CustomerName, LoadSize
FROM CHIP_DELIVERY
WHERE CustomerName IN
(SELECT OwnerName
FROM OWNER
WHERE OwnerName in
(SELECT OwnerName
FROM SERVICE
WHERE AmountBilled IS NOT NULL
AND AmountPaid IS NULL));
O. Same as M, but use join with the JOIN ON syntax.
SELECT OwnerName, Phone
FROM OWNER JOIN SERVICE ON OWNER.OwnerName = SERVICE.OwnerName
WHERE AmountBilled IS NOT NULL
AND AmountPaid IS NULL);
P. Same as N, but use join with the JOIN ON syntax.
SELECT CustomerName, LoadSize
FROM CHIP_DELIVERY JOIN (OWNER JOIN SERVICE ON OWNER.OwnerName =
SERVICE.OwnerName) ON CHIP_DELIVERY.CustomerName =
SERVICE.OwnerName
WHERE AmountBilled IS NOT NULL
AND AmountPaid IS NULL;
Q. List the names, email addresses, and registration dates of all customer registrations.
6-17
Chapter Six Introduction to Structured Query Language (SQL)
R. Show the names and phone numbers of all customers who have had a both a service
and a chip delivery.
SELECT OwnerName, Phone
FROM OWNER
WHERE OwnerName IN
(SELECT OwnerName
FROM SERVICE)
AND OwnerName IN
(SELECT CustomerName
FROM CHIP_DELIVERY);
S. Show the names and phone numbers of all customers who have had a chip delivery, but
not a service.
SELECT OwnerName, Phone
FROM OWNER
WHERE OwnerName NOT IN
(SELECT OwnerName
FROM SERVICE)
AND OwnerName IN
(SELECT CustomerName
FROM CHIP_DELIVERY);
6-18