SQL Part II Yong Choi School of Business

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

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(*) – Count all of HW Class – Review the Part table first

SQL Query to Count Records

Example 18 SELECT count(*) FROM Part WHERE Class="HW";

SQL Examples – Aggregate Functions • Example 19: Save as example 19 – Find the number of customers and the total of their balances. – Calculate total: sum(field name)

SQL Query to Count Records and Calculate a Total

Example 19 SELECT count(*), Sum(Balance) FROM Customer;

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 Customer. Count and Balances. Total. – Change column name using “AS” command

SQL Query to Perform Calculations and Rename Fields

Example 20 SELECT count(*) AS Customer. Count, Sum(Balance) AS Balance. Total FROM Customer;

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. – Use “IN” command for combining two queries

SQL Query with Subquery

Example 21 SELECT Order. Num FROM Order. Line WHERE Part. Num IN (SELECT Part. Num FROM Part WHERE Warehouse='3');

SQL Examples - Grouping • Use GROUP 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. Also, sort the query in ascending order using the rep number.

SQL Query to Group Records

Example 22 SELECT Rep. Num, Count(*) AS Num. Of. Customer, Avg(Balance) AS Avg. Balance FROM Customer GROUP BY Rep. Num

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 Num. Of. Customers and Average. Balance. – Use of “Having” command.

SQL Query to Restrict Groups

Example 23 SELECT Rep. Num, count(*) AS Num. Customer, Avg(Balance) AS Average. Balance FROM Customer GROUP BY Rep. Num HAVING Count(*)<4;

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.

SQL Query with ‘WHERE’ and ‘HAVING’ Clauses

Example 23 -1 SELECT Rep. Num, count(*) AS Num. Customer, Avg(Balance) AS Average. Balance FROM Customer WHERE Credit. Limit<10000 GROUP BY Rep. Num HAVING Count(*)<3;

Pine Valley Furniture Company data model (from Chapter 1, Figure 1 -3) 23

Figure 7 -1 Pine Valley Furniture Company Customer_T and Order_T tables with pointers from customers to their orders These tables are used in queries that follow Chapter 7 Copyright © 2014 Pearson Education, Inc. 24 24

Equi-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 Custumer. ID from both tables B/C it is a common field between two as PK and FK. 25

Processing Multiple Tables • Type of Joins (driven from Set Theory) – Equi-join – Natural join – Outer join: Left or Right – Union join – Self join • Each example in the textbook chapter 7 26

Example visualization of different join types with results returned in shaded area 27

SQL Examples – Joining Tables • Use of multiple 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. • Customer. Num, Customer. Name, Rep. Num, Last. Name, First. Name • Delete Rep. Num 20 from Customer table. Then, try again

Customer. Num Customer. Name Street City State Zip 148 Al's Appliance and Sport 2837 Greenway Fillmore FL 33336 Balance Credit. Limit $6, 550 $7, 500 Rep. Num 20 Rep. Num Last. Name First. Name Street City State Zip 20 Kaiser Valerie 624 Randall Grove FL 33321 29 Commission $20, 542. 50 Rate 0. 05

Example 24 SELECT Customer. Num, Customer. Name, Rep. Num, Last. Name, First. Name FROM Customer, Rep WHERE Rep. Num=Customer. Rep. Num;

SQL Query to Join Tables

Select both: Customer. Rep. Num and Rep. Num With: WHERE Rep. Num=Customer. Rep. Num

Include Rep. Num but no where statement: WHERE Rep. Num = Customer. Rep. Num

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.

Query to Restrict Records in Join

Example 25 SELECT Customer. Num, Customer. Name, Rep. Num, Last. Name, First. Name FROM Customer, Rep WHERE Rep. Num=Customer. Rep. Num AND Credit. Limit=10000;

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. – How many tables? – How many conditions?

Example 26 Query to Join Multiple Tables

Example 26 SELECT Orders. Order. Num, Orderdate, Customer. Num, Customer. Name, Part. Num, Description, Num. Ordered, Quoted. Price FROM Orders, Customer, Order. Line, Part WHERE Customer. Num=Orders. Customer. Num AND Orders. Order. Num=Order. Line. Order. Num AND Order. Line. Part. Num=Part. Num;

SQL Examples – Union • 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.

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

Example 27 SELECT Customer. Num, Customer. Name FROM Customer WHERE Rep. Num='35' UNION SELECT Customer. Num, Customer. Name FROM Customer, Orders WHERE Customer. Num=Orders. Customer. Num;

Three Basic Functions by SQL And Their Basic SQL Commands 1. Data definition (last topic) through the use of CREATE 2. Data manipulation (next topic) through INSERT, UPDATE, and DELETE 3. Data querying (we are done with this) through the use of SELECT AND MANY OTHERS, which is the basis for all SQL queries.

SQL - Data Manipulation • Possible with Access – UPDATE – INSERT – DELETE • Possible with enterprise level DBMS – COMMIT – ROLLBACK

SQL - Data Manipulation (con’t) • UPDATE command makes data entry corrections UPDATE Project SET Prjt. Locat = 'Bellaire', Dept. Num = 5 WHERE Prjt. Num = 10; UPDATE Employee SET Salary = Salary * 1. 1 WHERE Branch = 'Lincoln';

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

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

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

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

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 32543. She has not yet earned any commission, but her commission rate is 5%(0. 05).

Example 29 INSERT INTO Rep VALUES ('16', 'Rands', 'Shron', '826 Raymond', 'Altonville', 'FL', '32543', 0, 0. 05);

SQL Examples - Data Manipulation • Example 30: Save as example 30 – Delete any row in the Orderline table in which the part number is BV 06 – First, review the part number BV 06 (Order. Num 21617)

Example 30 DELETE * FROM Order. Line WHERE Part. Num='BV 06';

SQL Examples – Creating a New Table Using a Existing Table • Example 31: save as example 31 – Create a new table named Small. Cust, 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

Example 31 SQL Query to Create New Table

Example 31 SELECT * INTO Small. Cust FROM Customer WHERE Credit. Limit<=7500;

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;

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>;

SQL Example – Data Definition • Example 32: Save as example 32 – Create a table call “CSUB” that contains following fields: • • Emp. ID Number (vs. Number(9) or Num(9)) Last. Name Char(20) First. Name Char(20) Street Char(30) City Char(20) State Char(2) Phone Number

Example 32 (con’t) Using Access Create table CSUB (Emp. ID Number, Last. Name Char(20), First. Name Char(20), Street Char(30), City Char(20), State Char(2), Phone Number); • Insert following values: – – – – Emp. ID: 123456789 Last. Name: your lastname First. Name: your firstname Street: 9001 Stockdale Hgwy City: Bakersfield State: CA Phone: 6616656691

Example 32 INSERT INTO Employee VALUES ('987654321', 'Choi', 'Yong', '9001 Stockdale', 'Bakersfield', 'CA', '123456789');

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