Lecture10: Data Manipulation in SQL, Advanced SQL queries

Transcription

1 IS220 / IS422 : Database Fundamentals : Data Manipulation in SQL, Advanced SQL queries Ref. Chapter5 Prepared by L. Nouf Almujally & Aisha AlArfaj College of Computer and Information Sciences - Information Systems Dept. 1

2 The Process of Database Design Real World Domain Conceptual model (ERD) Relational Data Model Create schema (DDL) Load Data (DML) 2

3 Sample Data in Customer Table custno custname custst custcity age 1 C1 Olaya St Jeddah 20 2 C2 Mains St Riyadh 30 3 C3 Mains Rd Riyadh 25 4 C4 Mains Rd Dammam 5 C5 Mains Rd Riyadh 3

4 Sample Data in Product Table prodno prodname proddes price 100 P0 Food P1 healthy food P P3 self_raising flour,80%wheat P4 network 80x 300 4

5 Sample Data in Orders Table ordno orddate custno prodno quantity 1 01-jan jan jan jan jan mar

6 EMPLOYEE Employee No. First Name Last Name Dept Number Salary E1 Mandy Smith D E2 Daniel Hodges D E3 Shaskia Ramanthan D E4 Graham Burke D E5 Annie Nguyen D DEPARTMENT Dept Number Dept Name Location Mail Number D1 Computer Science Bundoora 39 D2 Information Science Bendigo 30 D3 Physics Bundoora 37 D4 Chemistry Bendigo 35 6

7 Sample Data Table in orders Customer ( Example 3) Table O_Id OrderDate OrderPrice Customer /11/ Nora /10/ Sara /09/ Nora /09/ Nora /08/ Yara /10/ Sara 7

8 Table names and Column names Table name can be prefixed with the owner name. eg, if table product is owned by user John, you can use SELECT * FROM John.product; Column names can be prefixed with table name, SELECT product.prodno FROM product; 8

9 Alias SQL aliases are used to temporarily rename a table or a column heading. Syntax for Columns SELECT column_name AS alias_name FROM table_name; Syntax for Tables SELECT column_name(s) FROM table_name [AS] alias_name; 9

10 Alias ( important note ) Columns Alias: For example, you might wish to know how much is the combined total salary of all employees whose salary is above $25,000 / year. SELECT SUM(salary) AS "Total Salary" FROM employees WHERE salary > 25000; In this example, we've aliased the sum(salary) field as "Total Salary". As a result, "Total Salary" will display as the field name when the result set is returned. Table Alias: SELECT o.orderid, o.orderdate FROM Orders AS o; 10

11 Exercise create table count_null ( a number, b number ); Output : insert into count_null values ( 1, 5); insert into count_null values ( null, 7); insert into count_null values ( null, null); insert into count_null values ( 8, 2); select count(a) as "count_a_not_null", count(b) as "count_b_not_null", count(*) as "count_all from count_null; 11

12 JOIN 12

13 JOIN Often two or more tables are needed at the same time to find all required data These tables must be "joined" together The formal JOIN basically, computes a new table from those to be joined the new table contains data in the matching rows of the individual tables. 13

14 Types of JOIN types of JOIN:. INNER JOIN: Return rows when there is at least one match in both tables LEFT JOIN: Return all rows from the left table, even if there are no matches in the right table RIGHT JOIN: Return all rows from the right table, even if there are no matches in the left table Full Outer Joins : retains rows that are unmatched in both the tables. NOTE: In all the above outer joins, the displayed unmatched columns are filled with NULLS. 14

15 Types of JOIN 15

16 SQL Examples of Joins ( 1) Simple Join SELECT E.firstName, E.lastName, D.deptName FROM EMPLOYEE E, DEPARTMENT D WHERE E.deptNumber = D.deptNumber; Employee No. First Name Last Name Dept Number Salary E1 Mandy Smith D E2 Daniel Hodges D E3 Shaskia Ramanthan D E4 Graham Burke D E5 Annie Nguyen D Dept Number Dept Name Location Mail Number D1 Computer Science Bundoora 39 D2 Information Science Bendigo 30 D3 Physics Bundoora 37 D4 Chemistry Bendigo 35 16

17 This is the result from the matching Employee N First Name Last Name Salary Dept Number Dept Number Dept Name Location ail Numbe E1 Mandy Smith D1 D1 Computer Science Bundoora 39 E2 Daniel Hodges D2 D2 Information Science Bendigo 30 E3 Shaskia Ramanthan D2 D2 Information Science Bundoora 37 E4 Graham Burke D1 D1 Computer Science Bundoora 39 E5 Annie Nguyen D1 D1 Computer Science Bundoora 39 This is the final result: FIRSTNAME LASTNAME DEPTNAME Mandy Smith computer science Annie Nguyen computer science Graham Burke Computer Science Shaskia Ramanthan Information Science Daniel Hodges Information Science 17

18 SQL Examples of Joins ( 2 ) Joining more than two tables SELECT E.firstName, E.lastName, P.projTitle FROM EMPLOYEE E, WORKS_ON W, PROJECT P WHERE E.employeeNo = W.employeeNo AND W.projNo = P.projNo; WORKS_ON Employee No. ProjNo E1 1 E4 1 E5 2 E2 3 E3 1 ProjNo Project Title 1 Project A 2 Project B 3 Project C PROJECT EMPLOYEE Employee No. First Name Last Name Dept Number Salary E1 Mandy Smith D E4 Graham Burke D E5 Annie Nguyen D E2 Daniel Hodges D E3 Shaskia Ramanthan D FIRSTNAME LASTNAME PROJTITLE Mandy Smith Project A Graham Burke Project A Annie Nguyen Project B Daniel Hodges Project C Shaskia Ramanthan Project A 18

19 SQL Examples of Joins ( 3 ) List customers (by customer number, name and address) who have ordered the product 100. SELECT c.custno, custname, custst, custcity FROM customer c, orders o WHERE c.custno=o.custno AND prodno=100; CUSTNO CUSTNAME CUSTST CUSTCITY C1 Olaya St Jeddah 2 C2 Mains St Riyadh 3 C3 Mains Rd Riyadh custn o custn ame custst custcity age 1 C1 Olaya St Jeddah 20 2 C2 Mains St Riyadh 30 3 C3 Mains Rd Riyadh 25 4 C4 Mains Rd Dammam 5 C5 Mains Rd Riyadh ordno orddate custno prodno quantity 1 01-jan jan jan jan jan mar

20 SQL Examples of Joins ( 4 ) Find the total price of the products ordered by customer 1. SELECT sum(price*quantity) FROM orders, product WHERE orders.prodno = product.prodno AND custno = 1; SUM(PRICE*QUANTITY) prod No prodn ame proddes price 100 P0 Food P1 healthy food P P3 self_raising flour,80%wheat P4 network 80x 300 ordno orddate custno prodn o 1 01-jan jan jan jan jan quant ity 6 06-mar

21 Outer Joins in Oracle SQL Put an (+) on the potentially deficient side, ie the side where nulls may be added The (+) operator is placed in the join condition next to the table that is allowed to have NULL values. Example (Left Outer Join) : List all customers, and the products ordered if they have ordered some products. SELECT c.custno, o.prodno, quantity FROM customer c, orders o WHERE c.custno = o.custno (+); Note: a table may be outer joined with only one other table. Which table column to use is important, eg, in above example, do not use o.custno in place of c.custno in the SELECT list. 21

22 1) Inner Join SQL Example SELECT Student_Name, Advisor_Name FROM Students, Advisors WHERE Students.Advisor_ID= Advisors.Advisor_ID; 22

23 2) Left Outer Join SQL Example SELECT Student_Name, Advisor_Name FROM Students, Advisors WHERE Students.Advisor_ID= Advisors.Advisor_ID (+); 23

24 3) Right Outer Join SQL Example SELECT Student_Name, Advisor_Name FROM Students, Advisors WHERE Students.Advisor_ID(+)= Advisors.Advisor_ID ; Student_Name Advisor_Name Student_1 advisor 1 Student_5 advisor 3 Student_7 advisor 3 Student_9 advisor 1 24 Student_10 advisor 3 null Advisor 5

25 4) Full Outer Join SQL Example SELECT Student_Name, Advisor_Name FROM Students, Advisors WHERE Students.Advisor_ID (+) = Advisors.Advisor_ID (+) ; 25

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27 Nested Queries (1) Query results are tables, which can also be queried. SELECT * FROM (SELECT prodno, sum(quantity) AS sum FROM orders GROUP BY prodno) WHERE sum>10; Equivalent to PRODNO SUM SELECT prodno, sum(quantity) as sum FROM orders GROUP BY prodno HAVING sum(quantity)>10; PRODNO SUM The inner query is referred to as a subquery 27

28 prodno prodname proddes price Nested Queries (2) 100 P0 Food P1 healthy food P P3 self_raising flour,80%wheat P4 network 80x 300 If the query result is a single value, it can be treated as a value, and be compared with other values. Subquery with equality ( <, >) : Example: Find products with price more than average SELECT prodno, price FROM product WHERE price > (SELECT AVG(price) FROM product); AVG(PRICE) 200 PRODNO PRICE

29 Subquery Employee No. First Name Last Name Dept Number Salary E1 Mandy Smith D E2 Daniel Hodges D E3 Shaskia Ramanthan D E4 Graham Burke D E5 Annie Nguyen D Subquery with equality (=) : Give a list with first and last names of employees who work in the department with mail number = 39 SELECT firstname, lastname FROM EMPLOYEE Dept Number Dept Name Location Mail Number D1 Computer Science Bundoora 39 D2 Information Science Bendigo 30 D3 Physics Bundoora 37 D4 Chemistry Bendigo 35 WHERE deptnumber =(SELECT deptnumber FROM DEPARTMENT WHERE mailnumber = 39); FIRSTNAME LASTNAME mandy smith graham burke Annie nguyen DEPTNUMBER D1 29

30 Subquery Employee No. First Name Last Name Dept Number Salary E1 Mandy Smith D E2 Daniel Hodges D E3 Shaskia Ramanthan D E4 Graham Burke D E5 Annie Nguyen D Subquery with aggregate function: Give a list with first, last names and salary of employees whose salary is greater than average salary for all employees. SELECT firstname, lastname, salary FROM EMPLOYEE Dept Number Dept Name Location Mail Number D1 Computer Science Bundoora 39 D2 Information Science Bendigo 30 D3 Physics Bundoora 37 D4 Chemistry Bendigo 35 WHERE salary > (SELECT avg(salary) FROM EMPLOYEE); AVG(SALARY) FIRSTNAME LASTNAME SALARY Shaskia Ramanthan Annie Nguyen

31 Subquery Employee No. First Name Last Name Dept Number Salary E1 Mandy Smith D E2 Daniel Hodges D E3 Shaskia Ramanthan D E4 Graham Burke D E5 Annie Nguyen D Nested Subquery (use of IN): Give a list with first, last names of employees whose departments are located in Bundoora. SELECT firstname, lastname FROM EMPLOYEE WHERE deptnumber IN (SELECT deptnumber FIRSTNAME LASTNAME Annie nguyen graham burke mandy smith Dept Number Dept Name Location Mail Number D1 Computer Science Bundoora 39 D2 Information Science Bendigo 30 D3 Physics Bundoora 37 D4 Chemistry Bendigo 35 FROM DEPARTMENT WHERE lo atio = Bundoora ; deptnumber D1 D3 31

32 Subquery List the products ordered by customers living in Riyadh. SELECT prodno FROM orders WHERE custno IN (SELECT custno FROM customer WHERE custcity = Riyadh ; - This query is equivalent to custno SELECT prodno FROM orders o, customer c WHERE o.custno =c.custno AND custcity = Riyadh'; PRODNO

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34 Queries using EXISTS or NOT EXISTS Queries using EXISTS Designed for use only with subqueries EXISTS return true if there exists at least one row in the result table returned by the subquery, it is false if the subquery returns an empty result table. Syntax SELECT column_name FROM table_name WHERE EXISTS NOT EXISTS ( subquery ); 34

35 Queries using EXISTS or NOT EXISTS Example SELECT firstname, lastname FROM EMPLOYEE E FIRSTNAME LASTNAME Shaskia Ramanthan Daniel Hodges WHERE EXISTS (SELECT * FROM DEPARTMENT D WHERE E.deptNumber = D.deptNumber AND D.location = Bendigo ); Employee No. First Name Last Name Dept Number Salary E1 Mandy Smith D E2 Daniel Hodges D E3 Shaskia Ramanthan D E4 Graham Burke D E5 Annie Nguyen D Dept Number Dept Name Location Mail Number D1 Computer Science Bundoora 39 D2 Information Science Bendigo 30 D3 Physics Bundoora 37 D4 Chemistry Bendigo 35 35

36 Example. EXISTS custno custname custst custcity age 1 C1 Olaya St Jeddah 20 2 C2 Mains St Riyadh 30 3 C3 Mains Rd Riyadh 25 4 C4 Mains Rd Dammam 5 C5 Mains Rd Riyadh Find all customers who have ordered some products. SELECT * FROM customer c WHERE exists (SELECT * FROM orders o WHERE o.custno =c.custno); ordno orddate custn o prodno 1 01-jan jan jan jan jan mar CUSTNO CUNAME CUSTST CUSTCITY AGE C1 Olaya St Jeddah 20 2 C2 Mains St Riyadh 30 3 C3 Mains Rd Riyadh 25 If the subquery is not empty, then the exists condition is true. quant ity 36

37 Example. NOT EXISTS custno custname custst custcity age 1 C1 Olaya St Jeddah 20 2 C2 Mains St Riyadh 30 3 C3 Mains Rd Riyadh 25 4 C4 Mains Rd Dammam Find all customers such that no order made by them has a quantity less than 2. SELECT * FROM customer c WHERE NOT EXISTS (SELECT * FROM orders o WHERE o.custno = c.custno AND quantity <2); CUSTNO CUTNAME CUSTST CUSTCITY AGE C5 Mains Rd Riyadh 4 C4 Mains Rd Dammam 3 C3 Mains Rd Riyadh 25 5 C5 Mains Rd Riyadh ordno orddate custno prodno quant ity 1 01-jan jan jan jan jan mar

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39 UNION The UNION operator is used to combine the result-set of two or more SELECT statements. Notice that each SELECT statement within the UNION must 1. have the same number of columns. 2. The columns must also have similar data types. 3. the columns in each SELECT statement must be in the same order. Combines the results of two SELECT statements into one result set, and then eliminates any duplicate rows from that result set. SQL UNION Syntax SELECT column_name(s) FROM table_name1 UNION SELECT column_name(s) FROM table_name2 39

40 UNION Note: The UNION operator selects only distinct values by default. To allow duplicate values, use UNION ALL. UNION ALL Combines the results of two SELECT statements into one result set. SQL UNION ALL Syntax SELECT column_name(s) FROM table_name1 UNION ALL SELECT column_name(s) FROM table_name2 40

41 UNION Example 1 "Employees_Norway" E_ID E_Name 01 Hansen, Ola 02 Svendson, Tove 03 Svendson, Stephen 04 Pettersen, Kari Employees_USA E_ID E_Name 01 Turner, Sally 02 Kent, Clark 03 Svendson, Stephen 04 Scott, Stephen list all the different employees in Norway and USA SELECT E_Name FROM Employees_Norway UNION SELECT E_Name FROM Employees_USA; E_Name Hansen, Ola Svendson, Tove Svendson, Stephen Pettersen, Kari Turner, Sally Kent, Clark Scott, Stephen 41

42 UNION Example 2 SELECT custno FROM customer WHERE custcity= Riyadh UNION SELECT custno FROM orders WHERE prodno=102; CUSTNO // union of the two queries custno custnam e custst custcity age 1 C1 Olaya St Jeddah 20 2 C2 Mains St Riyadh 30 3 C3 Mains Rd Riyadh 25 4 C4 Mains Rd Dammam 5 C5 Mains Rd Riyadh ordno orddate custno prodno quantity 1 01-jan jan jan jan jan mar

43 MINUS the MINUS operator returns only unique rows returned by the first query but not by the second. Takes the result set of one SELECT statement, and removes those rows that are also returned by a second SELECT statement. SQL MINUS Syntax SELECT column_name(s) FROM table_name1 MINUS SELECT column_name(s) FROM table_name2 ; 43

44 MINUS Example 1 Example: List the products that had never been ordered by customers SELECT prodno FROM product MINUS SELECT prodno FROM orders; //difference from the two queries PRODNO prodno prodname proddes price 100 P0 Food P1 healthy food P P3 self_raising flour,80%wheat P4 network 80x 300 ordno orddate custno prodno quantity 1 01-jan jan jan jan jan mar

45 INTERSECT The INTERSECT operator returns only those rows returned by both queries. Returns only those rows that are returned by each of two SELECT statements. SQL INTERSECT Syntax SELECT column_name(s) FROM table_name1 INTERSECT SELECT column_name(s) FROM table_name2 ; 45

46 INTERSECT Example: List the customers from Riyadh city who ordered product 102 SELECT custno FROM customer WHERE custcity= Riyadh' INTERSECT SELECT custno FROM orders WHERE prodno=102; CUSTNO // intersect of the two queries custn o custname custst custcity age 1 C1 Olaya St Jeddah 20 2 C2 Mains St Riyadh 30 3 C3 Mains Rd Riyadh 25 4 C4 Mains Rd Dammam 5 C5 Mains Rd Riyadh ordno orddate custno prodno quantit y 1 01-jan jan jan jan jan mar

47 Examples EMPLOYEE Employee No. First Name Last Name Dept Number Salary E1 Mandy Smith D E2 Daniel Hodges D E3 Shaskia Ramanthan D DEPENDENT Employee No. First Name Last Name Date of Birth E1 Joshua Smith 12-Jun-1998 E3 Jay Ramanthan 04-Jan-1996 E1 Jemima Smith 08-Sep

48 Examples SELECT employeeno, firstname, lastname FROM EMPLOYEE UNION SELECT employeeno, firstname, lastname FROM DEPENDENT; SELECT employeeno FROM EMPLOYEE INTERSECT SELECT employeeno FROM DEPENDENT 48

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50 EMPLOYEE Table Example1 EMPLOYEE_ID LAST_NAME FIRST_NAME JOB_ID MANAGER_ID SALARY COMM DEPARTMENT_ID 7369 SMITH JOHN NULL ALLEN KEVIN DOYLE JEAN NULL DENNIS LYNN NULL BAKER LESLIE NULL WARK CYNTHIA SELECT department_id, count(*), max(salary), min(salary) FROM employee GROUP BY department_id; DEPARTMENT_ID COUNT(*) MAX(SALARY) MIN(SALARY)

51 EMPLOYEE Table Example2 EMPLOYEE_ID LAST_NAME FIRST_NAME JOB_ID MANAGER_ID SALARY COMM DEPARTMENT_ID 7369 SMITH JOHN NULL ALLEN KEVIN DOYLE JEAN NULL DENNIS LYNN NULL BAKER LESLIE NULL WARK CYNTHIA SELECT Employee_ID, FIRST_NAME,DEPARTMENT_ID FROM employee WHERE salary=(select max(salary) FROM employee); Employee_ID FIRST_NAME DEPARTMENT_ID JEAN 30 51

52 EMPLOYEE Table Example3 EMPLOYEE_ID LAST_NAME FIRST_NAME JOB_ID MANAGER_ID SALARY COMM DEPARTMENT_ID 7369 SMITH JOHN NULL ALLEN KEVIN DOYLE JEAN NULL DENNIS LYNN NULL BAKER LESLIE NULL WARK CYNTHIA SELECT Employee_ID FROM employee WHERE department_id IN (SELECT department_id FROM department WHERE a e= SALES ; DEPARTMENT Department_ID Name Location_ID 10 ACCOUNTING RESEARCH SALES OPERATIONS 167 EMPLOYEE_ID

53 EMPLOYEE Table Example4 EMPLOYEE_ID LAST_NAME FIRST_NAME JOB_ID MANAGER_ID SALARY COMM DEPARTMENT_ID 7369 SMITH JOHN NULL ALLEN KEVIN DOYLE JEAN NULL DENNIS LYNN NULL BAKER LESLIE NULL WARK CYNTHIA SELECT name FROM department d WHERE NOT EXISTS (SELECT last_name FROM employee e WHERE d.department_id=e.department_id); DEPARTMENT Department_ID Name Location_ID 10 ACCOUNTING RESEARCH SALES OPERATIONS 167 NAME ACCOUNTING 53

54 EMPLOYEE Table Example4 LAST_NAME DEPARTMENT_ID Department_ID Name SMITH RESEARCH ALLEN SALES DOYLE SALES DENNIS SALES BAKER OPERATIONS WARK SALES ACCOUNTING Department Not exist 54

55 EMPLOYEE Table Example5 EMPLOYEE_ID LAST_NAME FIRST_NAME JOB_ID MANAGER_ID SALARY COMM DEPARTMENT_ID 7369 SMITH JOHN NULL ALLEN KEVIN DOYLE JEAN NULL DENNIS LYNN NULL BAKER LESLIE NULL WARK CYNTHIA LAST_NAME DEPARTMENT_ID NAME BAKER 40 OPERATIONS 10 ACCOUNTING DEPARTMENT Department_ID Name Location_ID 10 ACCOUNTING RESEARCH SALES OPERATIONS 167 SELECT last_name, d.department_id, d.name FROM employee e, department d WHERE e.department_id (+)= d.department_id AND d.department_id in (SELECT department_id FROM depart e t WHERE a e IN ACCOUNTING, OPERATIONS ; 55

56 EMPLOYEE Table Example5 LAST_NAME LAST_NAME DEPARTMENT_ID DEPARTMENT_ID Department_ID Department_ID Name Name LAST_NAME SMITH SMITH Department_ID Name RESEARCH RESEARCH BAKER ALLEN ALLEN OPERATIONS SALES SALES DOYLE DOYLE ACCOUNTING SALES SALES DENNIS DENNIS SALES SALES BAKER BAKER OPERATIONS OPERATIONS WARK WARK SALES SALES LAST_NAME DEPARTMENT_ID NAME ACCOUNTING ACCOUNTING BAKER 40 OPERATIONS 10 ACCOUNTING SELECT last_name, d.department_id, d.name FROM employee e, department d WHERE e.department_id (+)= d.department_id AND d.department_id in (SELECT department_id FROM depart e t WHERE a e IN ACCOUNTING, OPERATIONS ; 56

57 EMPLOYEE Table Example6 EMPLOYEE_ID LAST_NAME FIRST_NAME JOB_ID MANAGER_ID SALARY COMM DEPARTMENT_ID 7369 SMITH JOHN NULL ALLEN KEVIN DOYLE JEAN NULL DENNIS LYNN NULL BAKER LESLIE NULL WARK CYNTHIA SELECT employee_id, First_name, Last_name, Salary FROM employee WHERE last_name like D% ; EMPL OYEE_ID FIRST_NAME LAST_NAME SALARY JEAN DOYLE LYNN DENNIS

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59 References Data ase Systems: A Practical Approach to Design, Implementation and Management. Tho as Connolly, Carolyn Begg. 5 th Edition, Addison-Wesley,