1. SQL SeQueL -Structured Query Language SQL1 -1986 SQL2 - 1992 better support for Algebraic operations SQL3 - 1999 Post-Relational row and column types, stored procedures, triggers, references, large objects

2. SQL v. Access Query Builder ‘Standard’ Database language used in all commercial DBMS’s GUI query-builder in Access cannot build complex SQL SQL is a textural language - store, edit, generate There are jobs in it!

3. SQL DATA MANIPULATION (DML) - factbase QUERY SELECT UPDATE DELETE INSERT DATA DEFINITION (DDL) -schema CREATE, ALTER, DROP DATA CONTROL (DCL) - access control GRANT,REVOKE

4. PROJECT - the columns SELECT dname FROM dept; Computed columns SELECT ename,sal*12 FROM emp; Renamed columns SELECT ename, sal*12 as annualSalary FROM emp;

5. RESTRICT - the rows SELECT * FROM emp WHERE job=‘Analyst’; complex conditions: SELECT * FROM emp WHERE job=‘Analyst’ or dept=20;

6. RESTRICT and PROJECT in SQL: SELECT Mgr FROM Emp WHERE job=‘Analyst’ or deptno=10; may produce duplicate rows: SELECT DISTINCT Mgr FROM Emp WHERE job=‘Analyst’ or deptno=10;

7. FUNCTIONS STRINGS LIKE DATE SYSDATE.. STATISTICAL FUNCTIONS COUNT, AVG, MIN, MAX, SUM GENERATE AGGREGATE VALUES SELECT SUM(sal) FROM emp; shows total salary over all emps

8. GROUP BY OFTEN WANT AGGREGATE GROUPS OF ROWS SELECT deptno, sum(sal) FROM emp GROUP BY deptno; Here the rows are first grouped together with their common values, then any aggregation is done on each subgroup.

9. HAVING Restrict Groups shown: SELECT deptno, sum(sal) FROM emp GROUP BY deptno HAVING SUM(sal) > 10000; After the groups have been aggregated, these new rows can be further selected.

10. SUBQUERY List all departments employing analysts select distinct deptno from emp where job = ‘analyst’ List the names of departments employing analysts SELECT Dname FROM Dept WHERE Deptno IN ( select distinct deptno from emp where job = ‘analyst’);

11. No Subquery? MySQL does not support subqueries (until version 4.1) How can we do the same query without subqueries Select distinct dname From dept natural join emp Where job=‘analyst’;

12. UNION SELECT deptno FROM dept WHERE dname like ‘%ale%’ UNION SELECT deptno FROM emp WHERE sal > 3000;

13. PRODUCT SELECT * from dept,emp; GENERATES N * M ROWS SO GENERALLY TOO LARGE (4 * 14 = 56 rows on the emp data) but BASIS OF ALL JOINS

14. PRODUCT THEN RESTRICT SELECT * FROM Dept, emp WHERE dept.deptno = emp.deptno called an EQUI-JOIN Now only matching rows in both tables appear since every emp has a non NULL deptno, there will be as many rows in the join as there are rows in the emp table

15. Product, project, restrict Can use SQL-2 operators in Access or MySQL SELECT dname, ename FROM dept inner join emp on dept.deptno=emp.deptno; Combine with restriction: SELECT dname, ename FROM dept inner join emp on dept.deptno=emp.deptno where job=‘analyst’;

16. JOIN or SUBQUERY? SELECT dname FROM dept inner join emp on dept.deptno=emp.deptno where job=‘analyst’ often several ways to express same query Which is ‘better’? What is ‘better’?

17. Outer Join Inner join returns only matching rows We can get non-matching rows too A LEFT JOIN B includes all the non matching rows in A as well A RIGHT JOIN B includes all the non matching rows in B as well a FULL JOIN B includes non matching rows from both tables

18. MF 23 fred23 sue 20 joe30 alice 34 bill34 julie Select * from M,F 23 fred23 sue 23 fred30 alice 23 fred34 julie 20 joe23 sue 20 joe30 alice 20 joe34 julie 34 bill23 sue 34 bill30 alice 34 bill 34 julie

19. MF 23 fred23 sue 20 joe30 alice 34 bill34 julie Select * from M inner join F on M.age = F.age 23 fred23 sue 23 fred30 alice 23 fred34 julie 20 joe23 sue 20 joe30 alice 20 joe34 julie 34 bill23 sue 34 bill30 alice 34 bill 34 julie

20. MF 23 fred23 sue 20 joe30 alice 34 bill34 julie Select * from M left join on M.age = F.age 23 fred23 sue 34 bill 34 julie

21. MF 23 fred23 sue 20 joe30 alice 34 bill34 julie Select * from M right join on M.age = F.age 23 fred23 sue 34 bill 34 julie

22. MF 23 fred23 sue 20 joe30 alice 34 bill34 julie Select * from M full join on M.age = F.age 23 fred23 sue 34 bill 34 julie

23. MF 23 fred23 sue 20 joe30 alice 34 bill34 julie Select * from M, F on M.age > F.age 23 fred23 sue 23 fred30 alice 23 fred34 julie 20 joe23 sue 20 joe30 alice 20 joe34 julie 34 bill23 sue 34 bill30 alice 34 bill 34 julie

24. Inner Join (natural Join) Select * from M inner join F on M.age=F.age; Left (Outer) Join Select * from M left join F on M.age=F.age; Right (Outer) Join Select * from M right join F on M.age=F.age; Full (outer) Join Select * from M full join F on M.age=F.age;

25. Show all departments without employees Select dname from dept left join emp on dept.deptno = emp.deptno shows all the department names where ename IS NULL; shows only the departments without employees