1. SQL Data Manipulation II Chapter 5 CIS 458 Sungchul Hong

2. Logical Operator AND, OR, NOT AB A AND BA OR BNOT A FFFFT TFFTF FTFTT TTTTF

3. Logical Operator DeMorgan’s Law – NOT (A AND B)  – NOT (A OR B)  – NOT (A AND B AND C) 

4. Row Selection (WHERE clause) Comparison Range Set membership Pattern match Null

5. Comparison Search Condition SELECT staffNo, fName, lName, position, salary FROM Staff WHERE salary > 10000;

6. SQL Comparison Operators = equals <> is not equal to < is less than > is greater than <= is less than or equal to >= is greater than or equal to

7. Order of Evaluation An expression is evaluated left to right Subexpressions in brackets are evaluated first. NOTs are evaluated before ANDs and Ors ANDs are evaluated before ORs

8. Compound Comparison Search SELECT * FROM Branch WHERE city = ‘London’ OR city = ‘Glasgow’;

9. Range Search Condition BETWEEN / NOT BETWEEN SELET staffNo, fName, lName, position, salary FROM Staff WHERE salary BETWEEN 20000 AND 30000; SELET staffNo, fName, lName, position, salary FROM Staff WHERE salary >= 20000 AND salary <= 30000;

10. Set Membership Search Condition IN / NOT IN SELECT staffNO, fName, lName, positon FROM Staff WHERE position IN (‘Manager’, ‘Supervisor’); SELECT staffNO, fName, lName, positon FROM Staff WHERE position =‘Manager’ OR position = ‘Supervisor’;

11. Pattern Match Search LIKE / NOT LIKE –% (* in MS): wildcard eg) LIKE ‘H*’ –_: single character eg) LIKE ‘H___’ –If the search string can include the pattern-matching character itself, then –LIKE ’15#%’ ESCAPE ‘#’ –SELECT ownerNo, fName, lName, address, telNO –FROM PrivateOwner –WHERE address LIKE ‘*Glasgow*’;

12. NULL Search Condition IS NULL / IS NOT NULL SELECT clientNO, viewDate FROM Viewing WHERE propertyNO = ‘PG4’ AND comment IS NULL; *

13. Sorting Results ORDER BY column identifiers ( ASC or DESC) SELECT staffNO, fName, lName, salary FROM Staff ORDER BY salary DESC;

14. Multiple Column Ordering SELECT properyNO, type, rooms, rent FROM PropertyForRent ORDER BY type, rent DESC;

15. Aggregate Functions COUNT: numeric/non-numeric –COUNT(*): counts all the rows –DISTINCT: do not count duplicate values SUM: numeric AVG: numeric MIN: numeric/non-numeric MAX: numeric/non-numeric

16. COUNT SELECT COUNT(*) AS count FROM PropertyForRent WHERE rent > 350; SELECT COUNT (DISTINCT propertyNO) AS count FROM Viewing WHERE viewDate BETWEEN ‘1-May-01’ AND ’31- May-01’;

17. Count & SUM SELECT COUNT (StaffNo) AS count, SUM(salary) AS sum FROM Staff WHERE position = ‘Manager’;

18. MIN, MAX, and AVG SELECT MIN(salary), MAX(salary), AVG(salary) AS avg FROM Staff;

19. Grouping GROUP BY In SELECT list, only column names, aggregate functions, constants, and expression of previous combination. All column names in the SELECT list must appear in the GROUP BY clause unless the name is used only in an aggregate function. Each item in the SELECT list must be single- valued per group.

20. GROUP BY Find the number of staff working in each branch and the sum of their salaries. SELECT branchNO, COUNT(staffNo) AS count, SUM(salary) AS sum FROM Staff GROUP BY branchNO ORDER BY branchNO; SELECT branchNO, (SELECT COUNT (staffNo) AS count FROM Staff s WHERE s.branchNo = b.branchNO), (SELECT SUM(salary) AS sum FROM Staff s WHERE s.branchNo = b.branchNo) FROM Branch b ORDER BY branchNO;

21. GROUP BY The result of GROUP BY branchNOstaffNosalary B003 B005 B007 SG37 SG14 SG5 SL21 SL41 SA9 12000.00 18000.00 24000.00 30000.00 9000.00 COUNT(staffNo)SUM(salary) 354000.00 239000.00 1 9000.00

22. GROUP BY branchNOstaffNosalary B003 B005 B007 SG37 SG14 SG5 SL21 SL41 SA9 12000.00 18000.00 24000.00 30000.00 9000.00 branchNOstaffNosalary B003 B005 B007 SG37 SG14 SG5 SL21 SL41 SA9 12000.00 18000.00 24000.00 30000.00 9000.00 bs

23. GROUP BY & HAVING For each branch office with more than one member of staff, find the number of staff working in each branch and the sum of their salaries. SELECT branchNo, COUNT(staffNo) AS count, SUM (salary) AS sum From Staff GROUP BY branchNO HAVING COUNT(staffNO)>1 ORDER BY branchNO;

24. Subqueries Inner SELECT statement Subquery or nested query SELECT, INSERT, UPDATE, and DELETE A scalar subquery returns a single column and a single row. A row subquery returns multiple columns, but again only a single row. A table subquery returns on or more columns and multiple rows.

25. Subquery with Equality List the staff who work in the branch at ‘163 Main St’ SELECT staffNO, fName, lName, position FROM Staff WHERE branchNO = (SELECT branchNO FROM Branch WHERE street = ‘163 Main St’);

26. Subquery with an Aggregate Function List all staff whose salary is greater than the average salary, and show by how much their salary is greater than the average. SELECT staffNO, fName, lName, position, salary – (SELECT AVG (salary) FROM Staff) AS salDiff FROM Staff WHERE salary > (SELECT AVG(salary) FROM Staff);

27. Subquery Rules The ORDER BY clause may not be used in a subquery. The subquery SELECT list must consist of a single column name or expression, except for subqueries that use the keyword EXISTS. BY default, column names in a subquery refer to the table name in the FROM clause of the subquery. It is possible to refer to a table in a FROM clause of an outer query by qualifying the column name. When a subquery is one of the two operands involved in a comparison, the subquery must appear on the right- hand side of the comparison.

28. Nested subqueries: Use of IN List the properties that are handled by staff who work in the branch at ‘163 Main St’. SELECT propertyNo, street, city, postcode, type, rooms, rent FROM PropertyForRent WHERE staffNo IN (SELECT staffNO FROM Staff WHERE branchNO = (SELECT branchNO FROM Branch WHERE street = ‘163 Main St’));

29. Use of ANY/SOME Find all staff whose salary is larger than the salary of at least one member of staff at branch B003. SELECT staffNO, fName, lName, position, salary FROM Staff WHERE salary > SOME (SELECT salary FROM Staff WHERE branchNo = ‘B003’);

30. Use of ALL Find all staff whose salary is larger than the salary of every member of staff at branch B003. SELECT staffNO, fName, lName, position, salary FROM Staff WHERE salary > ALL (SELECT salary FROM Staff WHERE branchNo =‘B003’);

31. Multi-Table Queries List the names of all clients who have viewed a property along with any comment supplied. SELECT c.clientNo, fName, lName, proertyNo, comment FROM Client c, Viewing v WHERE c.clientNo = v.clientNo

32. Database Updates INSERT UPDATE DELETE

33. INSERT INSERT INTO Staff VALUES (‘SG16’, ‘Alan’, ‘Brown’, ‘Assistant’, ‘M’, DATE ‘1957-05-25’, 8300, ‘B003’)

34. UPDATE UPDATE Staff SET salary = salary*1.03

35. DELETE DELETE FROM Viewing WHERE propertyNo = ‘PG4’;