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

2. 2 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

3. 3 Running Example R1 S1 S2  Instances of the Sailors and Reserves relations in our examples.

4. 4 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

5. 5 Example Query SELECT S.sname FROM Sailors S, Reserves R WHERE S.sid=R.sid AND R.bid=103

6. 6 Comparisons in Oracle

7. 7 Conceptual Evaluation Strategy  Semantics of SQL query defined in terms of 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 must find more efficient strategies to compute same answers. SELECT S.sname FROM Sailors S, Reserves R WHERE S.sid=R.sid AND R.bid=103

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

9. 9 Example of Conceptual Evaluation SELECT S.sname FROM Sailors S, Reserves R WHERE S.sid=R.sid AND R.bid=103

10. 10 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

11. 11 Find sailors who’ve reserved at least one boat  Question: What is the output of this query ? SELECT S.sid FROM Sailors S, Reserves R WHERE S.sid=R.sid

12. 12 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

13. 13 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.  MEANING of Query:  Find sailors (age of sailor and two fields defined by expressions) 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’

14. 14 SQL Queries Using Set Operations

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  If we replace OR by AND as done below, what are the semantics? 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 B.color=‘green’) Find sid’s of sailors who’ve reserved a red and a green boat.

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  INTERSECT: compute intersection of two union-compatible sets of tuples.  What happens if change S.sid as S.sname ? 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’ 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  SQL/92 standard, but some systems don’t support INTERSECT !  What would query look like without INTERSECT?

20. 20 SELECT S.sid FROM Sailors S, Boats B, Reserves R WHERE S.sid=R.sid AND R.bid=B.bid A ND 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  Could query be written 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 B1.color=‘red’ ) AND ( S.sid=R2.sid AND R2.bid=B2.bid AND B2.color=‘green’)  Do we need Sailors relation in the FROM clauses?

21. 21 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 : compute Difference of 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'))

22. 22 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 instead? 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'))  SQL/92 standard, but some systems don’t support EXCEPT.  What then?

23. 23 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 instead? 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'))  Except is SQL/92 standard, but some systems don’t. support EXCEPT.  What then?  Is the above query correct, or not correct ?

24. 24 About duplication  Default: Eliminate duplicate  Unless we use special keyword “ALL” : # of copies of a row in result UNION ALL(m+n) INTERSECT ALLmin(m,n) EXCEPT ALLm-n

25. 25 Nested Queries

26. 26 Nested Queries  WHERE, FROM, HAVING clauses can itself contain another SQL query, called a subquery.  powerful feature of SQL. 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:

27. 27 Nested Queries Evaluation  Nested loops evaluation:  For each Sailors tuple, Evaluate the WHERE clause qualification Which here means re-compute 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:

28. 28 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)

29. 29 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:

30. 30 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:

31. 31 Nested Queries with Correlation  Why, in general, this subquery must be re- computed for each Sailors tuple ?  Could this query be rewritten into a flat SQL query? 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:

32. 32 Nested Queries with Correlation New Query :  Finds sailors with at most one reservation for boat #103.  UNIQUE is used instead of EXISTS UNIQUE means no duplicates in set, or set is NULL.  “*” must be replaced by R.bid. (Why?) Otherwise, result tuples with different dates for 103 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:

33. 33 More on Set-Comparison Operators  op ANY and op ALL where op = { }  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’)

34. 34 More on Set-Comparison Operators  What if there is no other sailor with that name?  “> ALL” would return TRUE if set is empty. SELECT * FROM Sailors S WHERE S.rating > ALL ( SELECT S2.rating FROM Sailors S2 WHERE S2.sname=‘Horatio’)  Find sailors whose rating is greater than that of all sailors called Horatio:

35. 35 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’)

36. 36 Rewriting INTERSECT Queries Using IN  INTERSECT queries re-written using IN.  Note: This query can return “sname” (and not just “sid) of Sailors who’ve reserved both red and green boats Find sname’s of sailors who’ve reserved both a red and a green boat: SELECT S.sname 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’)

37. 37 Division in SQL Find sailors who’ve reserved all boats.

38. 38 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.

39. 39 Division in SQL  Can we 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)) Find sailors who’ve reserved all boats. (1)

40. 40 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)

41. 41 Aggregation and Having Clauses

42. 42 Aggregate Operators  Significant extension of relational algebra. COUNT (*) COUNT ( [ DISTINCT ] A) SUM ( [ DISTINCT ] A) AVG ( [ DISTINCT ] A) MAX (A) MIN (A) Why no Distinct?

43. 43 Aggregate Operators 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’

44. 44 More Examples of Aggregate Operators 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)

45. 45 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’

46. 46 Find name and age of the oldest sailor(s)  The first query is illegal! (Why ?) 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  Second and third example queries equivalent ( SQL/92 )  Third NOT supported in some systems.

47. 47 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 ?

48. 48 Motivation for Grouping  So far, aggregate operators to all (qualifying) tuples.  Question?  What if want to apply aggregate to each group of tuples.  Example :  Find the age of the youngest sailor for each rating level.  Example Procedure :  Suppose rating values  {1,2,…,10}, 10 queries: SELECT MIN (S.age) FROM Sailors S WHERE S.rating = i For i = 1, 2,..., 10:

49. 49 Motivation for Grouping  What’s the problem with above ? We don’t know how many rating levels exist. Nor what the rating values for these levels are. SELECT MIN (S.age) FROM Sailors S WHERE S.rating = i For i = 1, 2,..., 10:

50. 50 Add Group By to SQL

51. 51 Queries With GROUP BY and HAVING  A group is a set of tuples that each have the same value for all attributes in grouping-list.  HAVING clause is a restriction on each group. SELECT [DISTINCT] target-list FROM relation-list WHERE qualification GROUP BY grouping-list HAVING group-qualification

52. 52 Queries With GROUP BY and HAVING  target-list contains : (i) attribute names (ii) aggregate-op (column-name) (e.g., MIN ( S.age )).  A group is a set of tuples that each have the same value for all attributes in grouping-list.  REQUIREMENT: - target-list (i)  grouping-list. - Why? - E ach answer tuple of a group must have single value. SELECT [DISTINCT] target-list FROM relation-list WHERE qualification GROUP BY grouping-list HAVING group-qualification

53. 53 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

54. 54 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.

55. 55 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 Sailors instance:

56. 56 Find age of the youngest sailor with age 18, for each rating with at least 2 such sailors.

57. 57 Find age of 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. Options: 1.Put 60 age condition into WHERE clause 2.Put 60 age condition into HAVING clause.

58. 58 Find age of 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)

59. 59 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) EVERY : Holds for all tuples in the group.

60. 60 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: Now check age<=60 before making groups.

61. 61 For each red boat, find the number of reservations for this boat  Grouping over Join of three relations.  Q: What do we get if we remove B.color=‘red’ from WHERE clause and add HAVING clause with this condition? No difference. But Illegal. Only column in GroupBy can appear in Having, unless in aggregate operator of Having SELECT B.bid, COUNT (*) AS s-count 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

62. 62 Find age of the youngest sailor with age > 18, for each rating with at least 2 sailors (of any age) 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. Find age of the youngest sailor with age > 18, for each rating with at least 2 such sailors (of age 18)

63. 63 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)

64. 64 Find those ratings for which the average age is the minimum over all ratings SELECT Temp.rating, Temp.avg-age FROM (SELECT S.rating, AVG (S.age) AS avg-age FROM Sailors S GROUP BY S.rating) AS Temp WHERE Temp.avg-age = (SELECT MIN (Temp.avg-age) FROM Temp)  Correct solution (in SQL/92):

65. 65 Null Values

66. 66 Null Values  Field values in a tuple are sometimes :  Unknown (e.g., a rating has not been assigned) or  Inapplicable (e.g., no maiden-name when a male person)  SQL provides a special value null for such situations.  The presence of null complicates many issues.

67. 67 Comparisons Using Null Values  (S.rating = 8) is TRUE or FALSE ??  What if rating has null value in tuple ?  We need a 3-valued logic : condition can be true, false or unknown.  SQL special operators IS NULL to check if value is/is not null.  Disallow NULL value : rating INTEGER NOT NULL

68. 68 Comparison with NULL values  Arithmetic operations on NULL return NULL.  Comparison operators on NULL return UNKNOWN.  We can explicitly check whether a value is null or not by IS NULL and by IS NOT NULL.

69. 69 Truth table with UNKNOWN UNKNOWN AND TRUE = UNKNOWN UNKNOWN OR TRUE = TRUE UNKNOWN AND FALSE = FALSE UNKNOWN OR FALSE = UNKNOWN UNKNOWN AND UNKNOWN = UNKNOWN UNKNOWN OR UNKNOWN = UNKNOWN NOT UNKNOWN = UNKNOWN WHERE clause is satisfied only when it evaluates to TRUE.

70. 70 Outer Joins : Special Operators  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 Sailors rows (left) without a matching Reserves row (right) appear in result, but not vice versa.

71. 71 Summary  SQL was important factor in early acceptance of relational model : easy-to-understand !  Relationally complete: in fact, significantly more expressive power than relational algebra.  Many alternative ways to write a query. So optimizer must look for most efficient evaluation plan.  In practice, users (still) need to be aware of how queries are optimized and evaluated for best results.