1. Yong Choi School of Business CSU, Bakersfield
SQL – Part II
Yong Choi
School of Business
CSU, Bakersfield

2. SQL Examples – Aggregate Functions
Example 18: Save as example 18
How many parts (count number of records) are in item class HW?
Use of “count” command
Count all records: count(*)
No idea? Try to figure out manually!

3. SQL Query to Count Records
Example 18
SQL Query to Count Records

4. Example 18
SELECT count(*)
FROM Part
WHERE Class=_____________;

5. SQL Examples – Aggregate Functions
Example 19: Save as example 19
Find the number of customers and the total of their balances.

6. SQL Query to Count Records and Calculate a Total
Example 19
SQL Query to Count Records and Calculate a Total

7. Example 19
SELECT count(*), Sum(____________)
FROM Customer;

8. SQL Examples – Aggregate Functions
Example 20: Save as example 20
Find the total number of customers and the total of their balances. Change the column names for the number of customers and the total of their balances to CustomerCount and BalanceTotal.
Assign column name using “AS” command

9. SQL Query to Perform Calculations and Rename Fields
Example 20
SQL Query to Perform Calculations and Rename Fields

10. Example 20 SELECT count(*) AS ___________,
Sum(Balance) AS _____________
FROM Customer;

11. SQL Examples – Nested Query
A query inside another query
A inside query (sub-query) is evaluated first. It is common to enclose sub-query in parentheses for readability!!
Example 21: Save as example 21
List the order number for each order from the order line table for a part located in warehouse 3. That is, we are looking for list of OrderNum that is associated with warehouse 3.
No idea? Try to figure out manually!

12. SQL Query with Subquery
Example 21
SQL Query with Subquery

13. Example 21 SELECT OrderNum FROM OrderLine WHERE PartNum IN

14. SQL Examples - Grouping
Use GROUP BY clause
ONLY grouping, NOT sorting (usually associated with ORDER BY clause)
Example 22: Save as example 22
For each sales rep, list the rep number, the number of customers assigned to each rep, and the average balance of the rep’s customers.
Rename the count of the number of customers and the average of the balances to NumOfCustomers and AverageBalance

15. # of customers are grouped by each sales rep #
Example 22
SQL Query to Group Records
# of customers are grouped by each sales rep #

16. Example 22
SELECT RepNum,
Count(_____) AS NumOfCustomer, Avg(Balance) AS AvgBalance
FROM Customer
GROUP BY ____________;

17. SQL Examples – Grouping (con’t)
Example 23: Save as example 23
For each sales rep with fewer than four customers, list the rep number, the number of customers assigned to the rep, and the average balance of the rep’s customers. Rename the count of the number of customers and the average of the balances to NumOfCustomers and AverageBalance.
Use of “Having” command.
Count, Rep. with < 4 customers (group)

18. SQL Query to Restrict Groups
Example 23
SQL Query to Restrict Groups
# of customers are grouped by each sales rep #
Apply the condition to the group

19. Example 23 where to put “Having” command? SELECT RepNum,
count(*) AS NumCustomer,
Avg(Balance) AS AverageBalance
FROM Customer
GROUP BY RepNum;

20. SQL Examples – Grouping (con’t)
Use of Where and Having clauses together
“Where” command must be stated first
Example 23-1: Save as example 23-1
Exactly same as example 23. Except, only groups with fewer than three records and customers with credit limit of less than $10,000 must be included.

21. Example 23-1
SQL Query with ‘WHERE’
and ‘HAVING’ Clauses

22. Example 23-1 Where to put “Where” command?
SELECT RepNum, count(*) AS NumCustomer, Avg(Balance) AS AverageBalance
FROM Customer
GROUP BY RepNum
HAVING Count(*)<3;

23. Exercise Question 8 from chapter 6
How many students are enrolled on Database and Networking? (Hint: use SectionNo for each class so you can determine the answer from the semester of 2008)?
Which courses were taught in the first semester of 2008 (I-2008) but not in the second semester of 2008 (II-2008)? (Note: There is no indication of the second semester of 2008 (II-2008))?

24. Matching Field Try below SQLs using Matching_DB on the class website
To identify missing “RepNum,” make sure to review each table first!
SQL 1
SELECT Customer.RepNum, CustomerName, Rep.RepNum, LastName, FirstName
FROM Customer, Rep;
SQL 2
FROM Customer, Rep
Where Customer.RepNum=Rep.RepNum;

25. Inner Join Example
For each customer who placed an order, what is the customer’s name and order number?
The best way to find out match customers with their orders is Including CustumerID from both tables

27. Different Type of SQL Joins
(INNER) JOIN: Returns records that have matching values in both tables
LEFT (OUTER) JOIN: Return all records from the left table, and the matched records from the right table
RIGHT (OUTER) JOIN: Return all records from the right table, and the matched records from the left table
FULL (OUTER) JOIN: Return all records when there is a match in either left or right table
Also, review SQL Joins in the textbook chapter 7

28. SQL Examples – Joining Tables
Example 24: Save as example 24
List the number and name of each customer together with the number, last name, and first name of the sales rep who represents the customer. If a customer does not have a Rep., then he/she must not be included.
That is, look for matching RepNum from both tables

29. SQL Query to Join Tables
Example 24
SQL Query to Join Tables

30. Example 24
SELECT CustomerNum, CustomerName, Rep.RepNum, LastName, FirstName
FROM Customer, Rep
Where _______.RepNum =_________.RepNum;
FROM Customer INNER JOIN [Rep] ON ________.RepNum=________.RepNum;

31. SQL Examples – Joining Tables (con’t)
Use of multiple tables with a compound condition
Example 25: Save as example 25
List the number and name of each customer whose credit limit is $10,000 together with number, last name, and first name of the sales rep who represents the customer.

32. Query to Restrict Records in Join
Example 25
Query to Restrict Records in Join

33. Example 25
SELECT CustomerNum, CustomerName, Rep.RepNum, LastName, FirstName
FROM Customer, Rep
WHERE Rep.RepNum=Customer.RepNum AND
_______________________;

34. SQL Examples – Joining Tables (con’t)
Example 26: Save as example
For every order, list the order number, order date, customer number, and customer name. In addition, for each order line within the order, list the part number, description, number ordered, and quoted price. Make sure that everything is matched. In other words, any “unmatched” records must not be included.

35. Query to Join Multiple Tables
Example 26
Query to Join Multiple Tables

36. Example 26
SELECT Orders.OrderNum, Orderdate, Customer.CustomerNum, CustomerName, Part.PartNum, Description, NumOrdered, QuotedPrice
FROM Orders, Customer, OrderLine, Part
WHERE Customer.CustomerNum=Orders.CustomerNum
AND Orders.OrderNum=OrderLine.OrderNum
AND OrderLine.PartNum=Part.PartNum;

37. SQL Examples – Union Example 27: Save as example 27
The union of two tables is a table containing all rows that are in either the first table, the second table, or both tables.
Two tables involved in union must have same structure.
Example 27: Save as example 27
List the number and name of all customers that are either represented by sales rep 35 OR that currently have orders on file, OR both.
Any “unmatched” records must not be included.

38. Example 27 SQL Query to Perform Union
Red: Currently have orders on file
Blue: Represented by sales rep 35
Green: Both

39. Example 27 SELECT CustomerNum, CustomerName FROM Customer
WHERE RepNum='35'
UNION
SELECT Customer.CustomerNum, CustomerName
FROM Customer, Orders
WHERE Customer.CustomerNum=Orders.CustomerNum;

40. Three Basic Functions by SQL And Their Basic SQL Commands
Data definition (last topic) through the use of CREATE
Data manipulation (next topic) through INSERT, UPDATE, and DELETE
Data querying (we are done with this) through the use of SELECT AND MANY OTHERS, which is the basis for all SQL queries.

41. SQL - Data Manipulation
Possible with Access
UPDATE
INSERT
DELETE
Only possible with enterprise level DBMS
COMMIT
ROLLBACK

42. SQL - Data Manipulation (con’t)
UPDATE command makes data entry corrections
UPDATE Project
SET PrjtLocat = 'Bellaire', DeptNum = 5
WHERE PrjtNum = 10;
UPDATE Employee
SET Salary = Salary * 1.1
WHERE Branch = 'Lincoln';

43. SQL - Data Manipulation (con’t)
INSERT command add new data to a table
INSERT INTO Employee (SSN, LastName, FirstName)
VALUES ('Richard', 'Marini', '43433');
DELETE command removes table row
DELETE FROM Employee
WHERE LastName = 'Brown';

44. SQL - Data Manipulation (con’t)
COMMIT command store data on the secondary memory permanently
ROLLBACK command restores database back to previous condition if COMMIT hasn’t been used

45. SQL Examples - Data Manipulation
Example 28: Save as example 28
Update the street address of customer 524 to 1445 Rivard
First, review the current street address of customer 524 (838 Ridgeland)

46. Example 28 UPDATE Customer SET Street = '1445 Rivard'
WHERE CustomerNum='524';

47. SQL Examples - Data Manipulation
Example 29: Save as example 29
Add a new sales rep to the Rep table. Her number is 16, her name is Sharon Rands, and her address is 826 Raymond, Altonville, FL She has not yet earned any commission, but her commission rate is 5%(0.05).
First, review the “Rep” table

48. Example 29 INSERT INTO Rep
VALUES ('16', 'Rands', 'Shron', '826 Raymond', 'Altonville', 'FL', '32543', 0, 0.05);

49. SQL Examples - Data Manipulation
Example 30: Save as example 30
Delete any row in the Orderline table in which the part number is BV06
First, review the part number BV06 (OrderNum21617)

50. Example 30
DELETE *
FROM OrderLine
WHERE PartNum='BV06';

51. SQL Examples – Creating a New Table Using a Existing Table
Example 31: save as example 31
Create a new table named SmallCust, consisting of all fields from the Customer table and those rows in which the credit limit is less than or equal to $7,500.
SELECT
INTO Name of table to create
FROM
WHERE

52. SQL Query to Create New Table
Example 31
SQL Query to Create New Table

53. Example 31 SELECT * INTO SmallCust FROM Customer
WHERE CreditLimit<=7500;

54. SQL - Data Definition I
Create a database structure to hold all the database tables; MS Access ONLY can create tables
Usually, only a DBA can create a new database structure
SQL syntax for creating a database structure:
CREATE SCHEMA AUTHORIZATION <creator>;
Example: CREATE SCHEMA AUTHORIZATION JONES;

55. SQL - Data Definition II
Specify a new relation by giving it a name and specifying each of its attributes.
Each attribute is given a name, a data type to specify its values, and some constraints on the attribute.
Syntax:
CREATE TABLE <table name>;

56. SQL Example – Data Definition
Example 32: Save as example 32
Create a table call “CSUB” that contains following fields:
EmpID Number (vs. Number(9) or Num(9))
LastName Char(20)
FirstName Char(20)
Street Char(30)
City Char(20)
State Char(2)
Phone Number

57. Example 32 (con’t) Using Access Insert following values:
Create table CSUB
(EmpID Number, LastName Char(20), FirstName Char(20), Street Char(30), City Char(20), State Char(2), Phone Number);
Insert following values:
EmpID:
LastName: your lastname
FirstName: your firstname
Street: 9001 Stockdale Hgwy
City: Bakersfield
State: CA
Phone:

58. Example 32 INSERT INTO CSUB
VALUES (' ', 'Choi', 'Yong', '9001 Stockdale', 'Bakersfield', 'CA', ' ');

59. Using Oracle CREATE TABLE EMPLOYEE (FNAME VARCHAR(15) NOT NULL,
LNAME VARCHAR(15) NOT NULL,
SSN CHAR(9) NOT NULL, BDATE DATE,
SEX CHAR,
SALARY DECIMAL(10,2),
SUPERSSN CHAR(9),
DEPTNO INT NOT NULL, PRIMARY KEY (SSN),
FOREIGN KEY (SUPERSSN) REFERENCES EMPLOYEE(SSN), FOREIGN KEY (DNO) REFERENCES DEPARTMENT(DNUMBER) );