1. 1 SQL: Structured Query Language (‘Sequel’) Chapter 5

2. 2 SQL 1. Query Language 2. Constraints 3. Triggers

3. 3 SQL Overview  Data Manipulation Language (DML) Queries, insert, delete, modify  Data Definition Language (DDL) Creation, deletion, modification tables and views Creating and deleting index (commercial DB)  Triggers and Advanced Integrity Constraints  Embedded and Dynamic SQL Calling SQL code using host language (C, etc. ) (Chapter 6)  Client-Server Execution and Remote Database Access (chapter 7)  Transaction Management (chapter 21)  Security  Advanced features OO, recursive queries, data mining, spatial data, text and XML

4. 4 Running Example R1 S1 S2  Instances of the Sailors and Reserves relations in our examples.

5. 5 Basic SQL Query  relation-list A list of relation names  target-list A list of attributes of relations in relation-list  qualification Comparisons (Attr op const or Attr1 op Attr2, where op is one of ) combined using AND, OR and NOT.  DISTINCT is optional keyword indicating that answer should not contain duplicates. Default is that duplicates are not eliminated! SELECT [DISTINCT] target-list FROM relation-list WHERE qualification

6. 6 Example Query SELECT S.sname FROM Sailors S, Reserves R WHERE S.sid=R.sid AND R.bid=103

7. 7 Comparisons in Oracle

8. 8 Example Query SELECT S.sname FROM Sailors S, Reserves R WHERE S.sid=R.sid AND R.bid=103

9. 9 Conceptual Evaluation Strategy  Semantics of SQL query defined in terms of following conceptual evaluation strategy: Compute cross-product of relation-list. Discard resulting tuples if they fail qualifications. Delete attributes that are not in target-list. If DISTINCT is specified, eliminate duplicate rows.  Probably the least efficient way!  Optimizer will find more efficient strategies to compute the same answers.

10. 10 R S SELECT S.sname FROM Sailors S, Reserves R WHERE S.sid=R.sid AND R.bid=103

11. 11 Example of Conceptual Evaluation SELECT S.sname FROM Sailors S, Reserves R WHERE S.sid=R.sid AND R.bid=103

12. 12 A Note on Range Variables  Needed only if same relation appears twice in FROM clause.  The previous query can also be written as: SELECT S.sname FROM Sailors S, Reserves R WHERE S.sid=R.sid AND bid=103 SELECT sname FROM Sailors, Reserves WHERE Sailors.sid=Reserves.sid AND bid=103 It is good style, however, to use range variables always! OR

13. 13 Find sailors who’ve reserved at least one boat  Questions: What is the exact output of this query ? Would adding DISTINCT to this query make a difference? What is the effect of replacing S.sid by S.sname in SELECT clause? Would adding DISTINCT to this variant of the query make a difference? SELECT S.sid FROM Sailors S, Reserves R WHERE S.sid=R.sid

14. 14 Expressions and Strings  AS and = are two ways to name fields in result.  LIKE is used for string matching. `_’ stands for any one character and `%’ stands for 0 or more arbitrary characters.  Find triples (of ages of sailors and two fields defined by expressions) for sailors whose names begin and end with B and contain at least three characters. SELECT S.age, age1=S.age-5, 2*S.age AS age2 FROM Sailors S WHERE S.sname LIKE ‘B_%B’

15. 15  UNION compute union of any two union-compatible sets of tuples SELECT S.sid FROM Sailors S, Reserves R, Boats B WHERE S.sid=R.sid AND R.bid=B.bid AND (B.color=‘red’ OR B.color=‘green’) SELECT S.sid FROM Sailors S, Boats B, Reserves R WHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘red’ UNION SELECT S.sid FROM Sailors S, Boats B, Reserves R WHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘green’ Find sid’s of sailors who’ve reserved a red or a green boat

16. 16  If we replace OR by AND in the first version, what do we get? SELECT S.sid FROM Sailors S, Boats B, Reserves R WHERE S.sid=R.sid AND R.bid=B.bid AND (B.color=‘red’ OR B.color=‘green’) SELECT S.sid FROM Sailors S, Boats B, Reserves R WHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘red’ UNION SELECT S.sid FROM Sailors S, Boats B, Reserves R WHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘green’ Find sid’s of sailors who’ve reserved a red or a green boat

17. 17 SELECT S.sid FROM Sailors S, Boats B, Reserves R WHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘red’ INTERSECT SELECT S.sid FROM Sailors S, Boats B, Reserves R WHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘green’ Find sid’s of sailors who’ve reserved a red and a green boat  INTERSECT : Can be used to compute the intersection of any two union-compatible sets of tuples.  What happens if change S.sid as S.sname???  SQL/92 standard, but some systems don’t support INTERSECT ? Key field!

18. 18 SELECT S.sid FROM Sailors S, Boats B, Reserves R WHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘red’ INTERSECT SELECT S.sid FROM Sailors S, Boats B, Reserves R WHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘green’ Find sid’s of sailors who’ve reserved a red and a green boat  What would query look like without INTERSECT? SELECT S.sid FROM Sailors S, Boats B1, Reserves R1, Boats B2, Reserves R2 WHERE S.sid=R1.sid AND R1.bid=B1.bid AND S.sid=R2.sid AND R2.bid=B2.bid AND (B1.color=‘red’ AND B2.color=‘green’) Key field!

19. 19 SELECT S.sid FROM Sailors S, Boats B, Reserves R WHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘red’ EXCEPT SELECT S.sid FROM Sailors S, Boats B, Reserves R WHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘green’ Find sid’s of sailors who’ve reserved a red not a green boat  EXCEPT : Can be used to compute the Difference of any two union-compatible sets of tuples.  SQL/92 standard, but some systems don’t support it.  What happens if change S.sid as S.sname??? SELECT S.sid FROM Sailors S, Boats B1, Reserves R1, Boats B2, Reserves R2 WHERE S.sid=R1.sid AND R1.bid=B1.bid AND S.sid=R2.sid AND R2.bid=B2.bid AND (B1.color='red' AND NOT( B2.color = 'green'))

20. 20 SELECT S.sid FROM Sailors S, Boats B, Reserves R WHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘red’ EXCEPT SELECT S.sid FROM Sailors S, Boats B, Reserves R WHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘green’ Find sid’s of sailors who’ve reserved a red and not a green boat  What if we want to compute the DIFFERENCE, called EXCEPT in SQL?  SQL/92 standard, but some systems don’t support EXCEPT. SELECT S.sid FROM Sailors S, Boats B1, Reserves R1, Boats B2, Reserves R2 WHERE S.sid=R1.sid AND R1.bid=B1.bid AND S.sid=R2.sid AND R2.bid=B2.bid AND (B1.color='red' AND NOT( B2.color = 'green'))

21. 21 About duplication  Default: Eliminate duplicate  Unless Use# of copies of a row in result UNION ALL(m+n) INTERSECT ALLmin(m,n) EXCEPT ALLm-n

22. 22 Nested Queries  WHERE, FROM, HAVING clauses can itself contain an SQL query!  powerful feature of SQL.  nested loops evaluation: For each Sailors tuple, check the qualification by re-computing the subquery. SELECT S.sname FROM Sailors S WHERE S.sid IN ( SELECT R.sid FROM Reserves R WHERE R.bid=103) Find names of sailors who’ve reserved boat #103:

23. 23 Nested Queries Evaluation  Nested loops evaluation:  For each Sailors tuple, check qualification by re-computing subquery. SELECT S.sname FROM Sailors S WHERE S.sid IN ( SELECT R.sid FROM Reserves R WHERE R.bid=103) Find names of sailors who’ve reserved boat #103:

24. 24 Nested Queries  Change query to find sailors who’ve not reserved #103: SELECT S.sname FROM Sailors S WHERE S.sid IN ( SELECT R.sid FROM Reserves R WHERE R.bid=103) Find names of sailors who’ve reserved boat #103: SELECT S.sname FROM Sailors S WHERE S.sid NOT IN ( SELECT R.sid FROM Reserves R WHERE R.bid=103)

25. 25 Nested Queries  Could this query be rewritten without nesting? SELECT S.sname FROM Sailors S WHERE S.sid IN ( SELECT R.sid FROM Reserves R WHERE R.bid=103) Find names of sailors who’ve reserved boat #103:

26. 26 Nested Queries with Correlation  EXISTS is another set comparison operator, like IN. SELECT S.sname FROM Sailors S WHERE EXISTS ( SELECT * FROM Reserves R WHERE R.bid=103 AND S.sid=R.sid) Find names of sailors who’ve reserved boat #103:

27. 27 Nested Queries with Correlation  Why, in general, this subquery must be re- computed for each Sailors tuple ? SELECT S.sname FROM Sailors S WHERE EXISTS ( SELECT * FROM Reserves R WHERE R.bid=103 AND S.sid=R.sid) Find names of sailors who’ve reserved boat #103:

28. 28 Nested Queries with Correlation  Question: Finds sailors with at most one reservation for boat #103.  UNIQUE is used instead of EXISTS  * is replaced by R.bid. (Why?) SELECT S.sname FROM Sailors S WHERE EXISTS ( SELECT * FROM Reserves R WHERE R.bid=103 AND S.sid=R.sid) Find names of sailors who’ve reserved boat #103:

29. 29 More on Set-Comparison Operators  op ANY, op ALL, op IN  Op = { }  e.g. Find sailors whose rating is greater than that of some sailor called Horatio:  IN is equal to =ANY  NOT IN is equal to <>ANY SELECT * FROM Sailors S WHERE S.rating > ANY ( SELECT S2.rating FROM Sailors S2 WHERE S2.sname=‘Horatio’)

30. 30 Rewriting INTERSECT Queries Using IN Find sid’s of sailors who’ve reserved both a red and a green boat: SELECT S.sid FROM Sailors S, Boats B, Reserves R WHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘red’ AND S.sid IN ( SELECT S2.sid FROM Sailors S2, Boats B2, Reserves R2 WHERE S2.sid=R2.sid AND R2.bid=B2.bid AND B2.color=‘green’)

31. 31 Rewriting INTERSECT Queries Using IN  Similarly, EXCEPT queries re-written using NOT IN. To find names (not sid ’s) of Sailors who’ve reserved both red and green boats Find sid’s of sailors who’ve reserved both a red and a green boat: SELECT S.sid FROM Sailors S, Boats B, Reserves R WHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘red’ AND S.sid IN ( SELECT S2.sid FROM Sailors S2, Boats B2, Reserves R2 WHERE S2.sid=R2.sid AND R2.bid=B2.bid AND B2.color=‘green’)

32. 32 Division in SQL SELECT S.sname FROM Sailors S WHERE NOT EXISTS (( SELECT B.bid FROM Boats B) EXCEPT ( SELECT R.bid FROM Reserves R WHERE R.sid=S.sid)) Find sailors who’ve reserved all boats.

33. 33 Division in SQL  Let’s do it without EXCEPT : SELECT S.sname FROM Sailors S WHERE NOT EXISTS (( SELECT B.bid FROM Boats B) EXCEPT ( SELECT R.bid FROM Reserves R WHERE R.sid=S.sid)) SELECT S.sname FROM Sailors S WHERE NOT EXISTS ( SELECT B.bid FROM Boats B WHERE NOT EXISTS ( SELECT R.bid FROM Reserves R WHERE R.bid=B.bid AND R.sid=S.sid)) Sailors S such that... there is no boat B without... a Reserves tuple showing S reserved B Find sailors who’ve reserved all boats. (1) (2)

34. 34 Division in SQL  Let’s do it the hard way, without EXCEPT : SELECT S.sname FROM Sailors S WHERE NOT EXISTS (( SELECT B.bid FROM Boats B) EXCEPT ( SELECT R.bid FROM Reserves R WHERE R.sid=S.sid)) SELECT S.sname FROM Sailors S WHERE NOT EXISTS ( SELECT B.bid FROM Boats B WHERE NOT EXISTS ( SELECT R.bid FROM Reserves R WHERE R.bid=B.bid AND R.sid=S.sid)) Sailors S such that... there is no boat B without... a Reserves tuple showing S reserved B Find sailors who’ve reserved all boats. (1) (2)

35. 35 Aggregate Operators  Significant extension of relational algebra. COUNT (*) COUNT ( [ DISTINCT ] A) SUM ( [ DISTINCT ] A) AVG ( [ DISTINCT ] A) MAX (A) MIN (A) SELECT AVG (S.age) FROM Sailors S WHERE S.rating=10 SELECT COUNT (*) FROM Sailors S Why no Distinct? SELECT COUNT ( DISTINCT S.rating) FROM Sailors S WHERE S.sname=‘Bob’

36. 36 Aggregate Operators  Significant extension of relational algebra. SELECT AVG ( DISTINCT S.age) FROM Sailors S WHERE S.rating=10 SELECT S.sname FROM Sailors S WHERE S.rating= ( SELECT MAX (S2.rating) FROM Sailors S2)

37. 37 Aggregate Operators  Significant extension of relational algebra. COUNT (*) COUNT ( [ DISTINCT ] A) SUM ( [ DISTINCT ] A) AVG ( [ DISTINCT ] A) MAX (A) MIN (A) SELECT AVG (S.age) FROM Sailors S WHERE S.rating=10 SELECT COUNT (*) FROM Sailors S SELECT AVG ( DISTINCT S.age) FROM Sailors S WHERE S.rating=10 SELECT S.sname FROM Sailors S WHERE S.rating= ( SELECT MAX (S2.rating) FROM Sailors S2) Why no Distinct? SELECT COUNT ( DISTINCT S.rating) FROM Sailors S WHERE S.sname=‘Bob’

38. 38 Find name and age of the oldest sailor(s)  The first query is illegal! (reason? when we discuss GROUP BY.)  The 3th = the 2th ( SQL/92)  3th NOT supported in some systems. SELECT S.sname, MAX (S.age) FROM Sailors S SELECT S.sname, S.age FROM Sailors S WHERE S.age = ( SELECT MAX (S2.age) FROM Sailors S2) SELECT S.sname, S.age FROM Sailors S WHERE ( SELECT MAX (S2.age) FROM Sailors S2) = S.age

39. 39 Find name and age of the oldest sailor(s) SELECT S.sname, MAX (S.age) FROM Sailors S SELECT S.sname, S.age FROM Sailors S WHERE S.age = ( SELECT MAX (S2.age) FROM Sailors S2) Is this first query legal? Is this second query legal? No! Why not ?

40. 40 Motivation for Grouping  So far, aggregate operators to all (qualifying) tuples.  Question? apply aggregate to each group of tuples.  Consider: Find the age of the youngest sailor for each rating level.  Suppose rating values  {1,2,…,10}, 10 queries:  Question? we don’t know how many rating levels exist ? what the rating values for these levels are! SELECT MIN (S.age) FROM Sailors S WHERE S.rating = i For i = 1, 2,..., 10:

41. 41 Queries With GROUP BY and HAVING  The target-list contains : (i) attribute names (ii) aggregate-op (column-name) (e.g., MIN ( S.age )).  A group is a set of tuples that have the same value for all attributes in grouping-list.  REQUIREMENT: - The target-list (i)  grouping-list. - Why? each answer tuple of a group must have a single value. SELECT [DISTINCT] target-list FROM relation-list WHERE qualification GROUP BY grouping-list HAVING group-qualification

42. 42 Find age of the youngest sailor with age 18, for each rating with at least 2 such sailors SELECT S.rating, MIN (S.age) AS min-age FROM Sailors S WHERE S.age >= 18 GROUP BY S.rating HAVING COUNT (*) > 1

43. 43 GroupBy --- Conceptual Evaluation 1. Compute the cross-product of relation-list 2. Discard tuples that fail qualification 3. Delete ` unnecessary’ fields 4. Partition the remaining tuples into groups by the value of attributes in grouping-list. (GroupBy) 5. Eliminate groups using the group-qualification (Having)  We will have a single value per group ! That is, one answer tuple is generated per qualifying group.

44. 44 Find age of the youngest sailor with age 18, for each rating with at least 2 such sailors SELECT S.rating, MIN (S.age) AS min-age FROM Sailors S WHERE S.age >= 18 GROUP BY S.rating HAVING COUNT (*) > 1 Answer relation: Q: Why not rating =9 ? Sailors instance:

45. 45 Find age of the youngest sailor with age 18, for each rating with at least 2 such sailors.

46. 46 Find age of the youngest sailor with age 18, for each rating with at least 2 such sailors SELECT S.rating, MIN (S.age) AS min-age FROM Sailors S WHERE S.age >= 18 GROUP BY S.rating HAVING COUNT (*) > 1 Find age of the youngest sailor with age 18, for each rating with at least 2 such sailors and with every sailor under 60. HAVING COUNT (*) > 1 AND EVERY (S.age <=60)

47. 47 Find age of the youngest sailor with age 18, for each rating with at least 2 such sailors and with every sailor under 60. HAVING COUNT (*) > 1 AND EVERY (S.age <=60) What is the result of changing EVERY to ANY?

48. 48 Find age of the youngest sailor with age 18, for each rating with at least 2 sailors between 18 and 60. SELECT S.rating, MIN (S.age) AS min-age FROM Sailors S WHERE S.age >= 18 AND S.age <= 60 GROUP BY S.rating HAVING COUNT (*) > 1 Answer relation: Sailors instance:

49. 49 For each red boat, find the number of reservations for this boat  Grouping over a Join of three relations.  Q: What do we get if we remove B.color=‘red’ from the WHERE clause and add a HAVING clause with this condition? No difference. Illegal. Only column in GroupBy can appear in Having, unless in aggregate operator of Having  Q: What if we drop Sailors and the condition involving S.sid? SELECT B.bid, COUNT (*) AS scount FROM Sailors S, Boats B, Reserves R WHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘red’ GROUP BY B.bid

50. 50 Find age of the youngest sailor with age > 18, for each rating with at least 2 sailors (of any age)  Q: find only ratings with 2 sailors over 18 ?  What if HAVING clause is replaced by: (two such sailors)  HAVING COUNT (*) >1 SELECT S.rating, MIN (S.age) FROM Sailors S WHERE S.age > 18 GROUP BY S.rating HAVING 1 < (SELECT COUNT (*) FROM Sailors S2 WHERE S.rating=S2.rating)  HAVING clause can also contain a subquery.

51. 51 Find those ratings for which the average age is the minimum over all ratings  WRONG : Aggregate operations cannot be nested! SELECT S.rating FROM Sailors S WHERE S.age = (SELECT MIN (AVG (S2.age)) FROM Sailors S2) SELECT Temp.rating, Temp.avgage FROM (SELECT S.rating, AVG (S.age) AS avgage FROM Sailors S GROUP BY S.rating) AS Temp WHERE Temp.avgage = (SELECT MIN (Temp.avgage) FROM Temp)  Correct solution (in SQL/92):

52. 52 Null Values  Field values in a tuple are sometimes :  unknown (e.g., a rating has not been assigned) or  inapplicable (e.g., no spouse’s name).  SQL provides a special value null for such situations.  The presence of null complicates many issues.  rating>8 ?? If rating = null ? What about AND, OR and NOT connectives?

53. 53 Null Values  SQL provides a special value null  The presence of null complicates many issues.  rating>8 ?? If rating = null ? What about AND, OR and NOT connectives?  We need a 3-valued logic (true, false and unknown ).  Special operators IS NULL to check if value is/is not null. Disallow NULL value : NOT NULL  e.g., WHERE clause eliminates rows that don’t evaluate to true. Duplicates??? (treated as true implicitly)  problem!  New operators ( outer Joins ).

54. 54 Outer Joins  Left Outer Join  Right Outer Join  Full Outer Join SELECT S.sid, R.bid FROM Sailors S NATURAL LEFT OUTER JOIN Reserves R SELECT S.sid, R.bid FROM Sailors S LEFT OUTER JOIN Reserves R WHERE S.sid = R.sid Q: Right Outer Join? Full Outer Join?

55. 55 Summary  SQL was an important factor in the early acceptance of the relational model; more natural than earlier procedural query languages.  Relationally complete; in fact, significantly more expressive power than relational algebra.  Even queries that can be expressed in RA can often be expressed more naturally in SQL.  Many alternative ways to write a query; optimizer should look for most efficient evaluation plan.  In practice, users need to be aware of how queries are optimized and evaluated for best results.