1. Matthew P. Johnson, OCL3, CISDD CUNY, June 20051 OCL3 Oracle 10g: SQL & PL/SQL Session #4 Matthew P. Johnson CISDD, CUNY June, 2005

2. Matthew P. Johnson, OCL3, CISDD CUNY, June 2005 2 Agenda Go over labs 1 and 2 More SQL Lab 4

3. Matthew P. Johnson, OCL3, CISDD CUNY, June 2005 3 Review Examples from sqlzoo.netsqlzoo.net SELECT L FROM R 1, …, R n WHERE C SELECT L FROM R 1, …, R n WHERE C  L (  C (R 1 x … R n )

4. Matthew P. Johnson, OCL3, CISDD CUNY, June 2005 4 SQL e.g. Acc(name,ssn,balance) Q: Who has the largest balance? Conceptually:  name (Acc) -  a2.name (  a2.bal < Acc.bal (Acc x  a2 (Acc))) In SQL?

5. Matthew P. Johnson, OCL3, CISDD CUNY, June 2005 5 New topic: Subqueries Powerful feature of SQL: one clause can contain other SQL queries  Anywhere where a value or relation is allowed Several ways:  Selection  single constant (scalar) in SELECT  Selection  single constant (scalar) in WHERE  Selection  relation in WHERE  Selection  relation in FROM

6. Matthew P. Johnson, OCL3, CISDD CUNY, June 2005 6 Subquery motivation Consider standard multi-table example:  Purchase(prodname, buyerssn, etc.)  Person(name, ssn, etc.)  What did Christo buy? As usual, need to AND on equality identifying ssn’s row and buyerssn’s row SELECT Purchase.prodname FROM Purchase, Person WHERE buyerssn = ssn AND name = 'Christo'

7. Matthew P. Johnson, OCL3, CISDD CUNY, June 2005 7 Subquery motivation Purchase(prodname, buyerssn, etc.) Person(name, ssn, etc.) What did Conrad buy? Natural intuition:  Go find Conrad’s ssn  Then find purchases SELECT ssn FROM Person WHERE name = 'Christo' SELECT Purchase.prodname FROM Purchase WHERE buyerssn = Christo’s-ssn

8. Matthew P. Johnson, OCL3, CISDD CUNY, June 2005 8 Subqueries Subquery: copy in Conrad’s selection for his ssn: The subquery returns one value, so the = is valid If it returns more (or fewer), we get a run-time error SELECT Purchase.prodname FROM Purchase WHERE buyerssn = (SELECT ssn FROM Person WHERE name = 'Christo') SELECT Purchase.prodname FROM Purchase WHERE buyerssn = (SELECT ssn FROM Person WHERE name = 'Christo')

9. Matthew P. Johnson, OCL3, CISDD CUNY, June 2005 9 Operators on subqueries Several new operators applied to (unary) selections: 1. IN R 2. EXISTS R 3. UNIQUE R 4. s > ALL R 5. s > ANY R 6. x IN R > is just an example op Each expression can be negated with NOT

10. Matthew P. Johnson, OCL3, CISDD CUNY, June 2005 10 Subqueries with IN Product(name,maker), Person(name,ssn), Purchase(buyerssn,product) Q: Find companies Martha bought products from Strategy: 1. Find Martha’s ssn 2. Find products listed with that ssn as buyer 3. Find company names of those products SELECT DISTINCT Product.maker FROM Product WHERE Product.name IN (SELECT Purchase.product FROM Purchase WHERE Purchase.buyerssn = (SELECT ssn FROM Person WHERE name = 'Martha')) SELECT DISTINCT Product.maker FROM Product WHERE Product.name IN (SELECT Purchase.product FROM Purchase WHERE Purchase.buyerssn = (SELECT ssn FROM Person WHERE name = 'Martha'))

11. Matthew P. Johnson, OCL3, CISDD CUNY, June 2005 11 Subqueries returning relations Equivalent to: SELECT DISTINCT Product.maker FROM Product, Purchase, People WHERE Product.name = Purchase.product AND Purchase.buyerssn = ssn AND name = 'Martha' SELECT DISTINCT Product.maker FROM Product, Purchase, People WHERE Product.name = Purchase.product AND Purchase.buyerssn = ssn AND name = 'Martha'

12. Matthew P. Johnson, OCL3, CISDD CUNY, June 2005 12 FROM subqueries Motivation for another way:  suppose we’re given Martha’s purchases  Then could just cross with Products to get product makers  Substitute (named) subquery for Martha’s purchases SELECT Product.maker FROM Product, (SELECT Purchase.product FROM Purchase WHERE Purchase.buyerssn = (SELECT ssn FROM Person WHERE name = 'Martha')) Marthas WHERE Product.name = Marthas.product SELECT Product.maker FROM Product, (SELECT Purchase.product FROM Purchase WHERE Purchase.buyerssn = (SELECT ssn FROM Person WHERE name = 'Martha')) Marthas WHERE Product.name = Marthas.product

13. Matthew P. Johnson, OCL3, CISDD CUNY, June 2005 13 ALL op Employees(name, job, divid, salary) Find which employees are paid more than all the programmers SELECT name FROM Employees WHERE salary > ALL (SELECT salary FROM Employees WHERE job='programmer') SELECT name FROM Employees WHERE salary > ALL (SELECT salary FROM Employees WHERE job='programmer')

14. Matthew P. Johnson, OCL3, CISDD CUNY, June 2005 14 ANY/SOME op Employees(name, job, divid, salary) Find which employees are paid more than at least one vice president SELECT name FROM Employees WHERE salary > ANY (SELECT salary FROM Employees WHERE job='VP') SELECT name FROM Employees WHERE salary > ANY (SELECT salary FROM Employees WHERE job='VP')

15. Matthew P. Johnson, OCL3, CISDD CUNY, June 2005 15 ANY/SOME op Employees(name, job, divid, salary) Find which employees are paid more than at least one vice president SELECT name FROM Employees WHERE salary > SOME (SELECT salary FROM Employees WHERE job='VP') SELECT name FROM Employees WHERE salary > SOME (SELECT salary FROM Employees WHERE job='VP')

16. Matthew P. Johnson, OCL3, CISDD CUNY, June 2005 16 Existential/Universal Conditions Employees(name, job, divid, salary) Division(name, id, head) Find all divisions with an employee whose salary is > 100000 Existential: easy! SELECT DISTINCT Division.name FROM Employees, Division WHERE salary > 100000 AND divid=id SELECT DISTINCT Division.name FROM Employees, Division WHERE salary > 100000 AND divid=id

17. Matthew P. Johnson, OCL3, CISDD CUNY, June 2005 17 Existential/Universal Conditions Employees(name, job, divid, salary) Division(name, id, head) Find all divisions in which everyone makes > 100000 Universal: hard!

18. Matthew P. Johnson, OCL3, CISDD CUNY, June 2005 18 Existential/universal with IN 2. Select the divisions we didn’t find 1. Find the other divisions: in which someone makes <= 100000 SELECT name FROM Division WHERE id IN (SELECT divid FROM Employees WHERE salary <= 100000 SELECT name FROM Division WHERE id IN (SELECT divid FROM Employees WHERE salary <= 100000 SELECT name FROM Division WHERE id NOT IN (SELECT divid FROM Employees WHERE salary <= 100000 SELECT name FROM Division WHERE id NOT IN (SELECT divid FROM Employees WHERE salary <= 100000

19. Matthew P. Johnson, OCL3, CISDD CUNY, June 2005 19 Acc(name,bal,type…) Q: Who has the largest balance? Can we do this with subqueries?

20. Matthew P. Johnson, OCL3, CISDD CUNY, June 2005 20 Last time: Acc(name,bal,type,…) Q: Find holder of largest account SELECT name FROM Acc WHERE bal >= ALL (SELECT bal FROM Acc) SELECT name FROM Acc WHERE bal >= ALL (SELECT bal FROM Acc) Correlated Queries

21. Matthew P. Johnson, OCL3, CISDD CUNY, June 2005 21 Correlated Queries So far, subquery executed once;  result used for higher query More complicated: correlated queries “[T]he subquery… [is] evaluated many times, once for each assignment of a value to some term in the subquery that comes from a tuple variable outside the subquery” (Ullman, p286). Q: What does this mean? A: That subqueries refer to vars from outer queries

22. Matthew P. Johnson, OCL3, CISDD CUNY, June 2005 22 Last time: Acc(name,bal,type,…) Q2: Find holder of largest account of each type SELECT name, type FROM Acc WHERE bal >= ALL (SELECT bal FROM Acc WHERE type=type) SELECT name, type FROM Acc WHERE bal >= ALL (SELECT bal FROM Acc WHERE type=type) Correlated Queries correlation

23. Matthew P. Johnson, OCL3, CISDD CUNY, June 2005 23 Last time: Acc(name,bal,type,…) Q2: Find holder of largest account of each type Note: 1. scope of variables 2. this can still be expressed as single SFW SELECT name, type FROM Acc a1 WHERE bal >= ALL (SELECT bal FROM Acc WHERE type=a1.type) SELECT name, type FROM Acc a1 WHERE bal >= ALL (SELECT bal FROM Acc WHERE type=a1.type) Correlated Queries correlation

24. Matthew P. Johnson, OCL3, CISDD CUNY, June 2005 24 EXCEPT and INTERSECT (SELECT R.A, R.B FROM R) INTERSECT (SELECT S.A, S.B FROM S) (SELECT R.A, R.B FROM R) INTERSECT (SELECT S.A, S.B FROM S) (SELECT R.A, R.B FROM R) EXCEPT (SELECT S.A, S.B FROM S) (SELECT R.A, R.B FROM R) EXCEPT (SELECT S.A, S.B FROM S) SELECT R.A, R.B FROM R WHERE EXISTS(SELECT * FROM S WHERE R.A=S.A and R.B=S.B) SELECT R.A, R.B FROM R WHERE EXISTS(SELECT * FROM S WHERE R.A=S.A and R.B=S.B) SELECT R.A, R.B FROM R WHERE NOT EXISTS(SELECT * FROM S WHERE R.A=S.A and R.B=S.B) SELECT R.A, R.B FROM R WHERE NOT EXISTS(SELECT * FROM S WHERE R.A=S.A and R.B=S.B)

25. Matthew P. Johnson, OCL3, CISDD CUNY, June 2005 25 More on Set-Comparison Operators We’ve already seen IN R, NOT IN R. Can also use EXISTS R, NOT EXISTS R Also available: op ANY R, op ALL R Find sailors whose rating is greater than that of some sailor called Alberto: SELECT R.SID FROM Reserves R WHERE R.rating > ANY ( SELECT R2.rating FROM Reserves R2 WHERE R2.sname=‘Alberto’)

26. Matthew P. Johnson, OCL3, CISDD CUNY, June 2005 26 Extended e.g. Scenario: 1. Purchase(pid, seller-ssn, buyer-ssn, etc.) 2. Person(ssn, name, etc.) 3. Product(pid, name, etc.) Q: Who (give names) bought gizmos from Dick? Where to start? Purchase uses pid, ssn, so must get them…

27. Matthew P. Johnson, OCL3, CISDD CUNY, June 2005 27 Last time: Complex RA Expressions Scenario: 1. Purchase(pid, seller-ssn, buyer-ssn, etc.) 2. Person(ssn, name, etc.) 3. Product(pid, name, etc.) Q: Who (give names) bought gizmos from Dick? Where to start? Purchase uses pid, ssn, so must get them…

28. Matthew P. Johnson, OCL3, CISDD CUNY, June 2005 28 Complex RA Expressions Person Purchase Person Product  name='Dick'  name='Gizmo'  pid  ssn seller-ssn=ssnpid=pidbuyer-ssn=Person.ssn  name

29. Matthew P. Johnson, OCL3, CISDD CUNY, June 2005 29 Translation to SQL We’re converting the tree on the last slide into SQL The result of the query should be the names indicated above One step at a time, we’ll make the query more complete, until we’ve translated the English-language description to an actual SQL query We’ll also simplify the query when possible (the names of the people who bought gadgets from Dick)

30. Matthew P. Johnson, OCL3, CISDD CUNY, June 2005 30 Translation to SQL Blue type = actual SQL Black italics = description of subquery Note: the subquery above consists of purchase records, except with the info describing the buyers attached  In the results, the column header for name will be 'buyer' SELECT DISTINCT name buyer FROM (the info, along with buyer names, for purchases of gadgets sold by Dick) SELECT DISTINCT name buyer FROM (the info, along with buyer names, for purchases of gadgets sold by Dick)

31. Matthew P. Johnson, OCL3, CISDD CUNY, June 2005 31 Translation to SQL Note: the subquery in this version is being given the name P2 We’re pairing our rows from Person with rows from P2 SELECT DISTINCT name buyer FROM (SELECT * FROM Person, (the purchases of gadgets from Dick) P2 WHERE Person.ssn = P2.buyer-ssn) SELECT DISTINCT name buyer FROM (SELECT * FROM Person, (the purchases of gadgets from Dick) P2 WHERE Person.ssn = P2.buyer-ssn)

32. Matthew P. Johnson, OCL3, CISDD CUNY, June 2005 32 Translation to SQL We simplified by combining the two SELECTs SELECT DISTINCT name buyer FROM Person, (the purchases of gadgets from Dick) P2 WHERE Person.ssn = P2.buyer-ssn SELECT DISTINCT name buyer FROM Person, (the purchases of gadgets from Dick) P2 WHERE Person.ssn = P2.buyer-ssn

33. Matthew P. Johnson, OCL3, CISDD CUNY, June 2005 33 Translation to SQL P2 is still the name of the subquery It’s just been filled in with a query that contains two subqueries Outer parentheses are bolded for clarity SELECT DISTINCT name buyer FROM Person, (SELECT * FROM Purchases WHERE seller-ssn = (Dick’s ssn) AND pid = (the id of gadget)) P2 WHERE Person.ssn = P2.buyer-ssn SELECT DISTINCT name buyer FROM Person, (SELECT * FROM Purchases WHERE seller-ssn = (Dick’s ssn) AND pid = (the id of gadget)) P2 WHERE Person.ssn = P2.buyer-ssn

34. Matthew P. Johnson, OCL3, CISDD CUNY, June 2005 34 Translation to SQL Now the subquery to find Dick’s ssn is filled in SELECT DISTINCT name buyer FROM Person, (SELECT * FROM Purchases WHERE seller-ssn = (SELECT ssn FROM Person WHERE name='Dick') AND pid = (the id of gadget)) P2 WHERE Person.ssn = P2.buyer-ssn SELECT DISTINCT name buyer FROM Person, (SELECT * FROM Purchases WHERE seller-ssn = (SELECT ssn FROM Person WHERE name='Dick') AND pid = (the id of gadget)) P2 WHERE Person.ssn = P2.buyer-ssn

35. Matthew P. Johnson, OCL3, CISDD CUNY, June 2005 35 Translation to SQL And now the subquery to find Gadget’s product id is filled in, too Note: the SQL simplified by using subqueries  Not used in relational algebra SELECT DISTINCT name buyer FROM Person, (SELECT * FROM Purchases WHERE seller-ssn = (SELECT ssn FROM Person WHERE name='Dick') AND pid = (SELECT pid FROM Product WHERE name='Gadget')) P2 WHERE Person.ssn = P2.buyer-ssn SELECT DISTINCT name buyer FROM Person, (SELECT * FROM Purchases WHERE seller-ssn = (SELECT ssn FROM Person WHERE name='Dick') AND pid = (SELECT pid FROM Product WHERE name='Gadget')) P2 WHERE Person.ssn = P2.buyer-ssn