1. Introduction to Database Systems 1 SQL: The Query Language Relation Model : Topic 4

2. Introduction to Database Systems 2 Basic SQL Query v SELECT [DISTINCT] target-list v FROM relation-list v WHERE qualification v ORDER BY subset-of-target-list

3. Introduction to Database Systems 3 Conceptual Evaluation Strategy v Semantics defined by a conceptual evaluation strategy: –Compute the 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. –If ORDER BY is specified, sort the result.

4. Introduction to Database Systems 4 Example of Conceptual Evaluation SELECT S.sname FROM Sailors S, Reserves R WHERE S.sid=R.sid AND R.bid=103

5. Introduction to Database Systems 5 A Note on Range Variables v Really needed only if the same relation appears twice in the FROM clause. 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

6. Introduction to Database Systems 6 In alphabetical order, print names of sailors who’ve reserved a red 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’ ORDER BY S.sname v It is illegal to replace S.sname by S.sid in the ORDER BY clause! (Why?)

7. Introduction to Database Systems 7 Find sailors who’ve reserved at least one boat SELECT S.sid FROM Sailors S, Reserves R WHERE S.sid=R.sid v Would adding DISTINCT to this query make a difference? v What is the effect of replacing S.sid by S.sname in the SELECT clause? Would adding DISTINCT to this variant of the query make a difference?

8. Introduction to Database Systems 8 Expressions and Strings v Arithmetic expressions, string pattern matching: 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’

9. Introduction to Database Systems 9 Find sid’s of sailors who’ve reserved a red or 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’ 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’ v UNION : computes the union of any two union- compatible sets of tuples (which are themselves the result of SQL queries). v Also available: EXCEPT (i.e. “set-minus”). What do we get if we replace UNION by EXCEPT?

10. Introduction to Database Systems 10 Find sid’s of sailors who’ve reserved a red and a green boat 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’) 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’ Key field! v INTERSECT: computes the intersection of any two union-compatible sets of tuples. v Contrast symmetry of the UNION and INTERSECT queries with how much the other versions differ!

11. Introduction to Database Systems 11 Nested Queries 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: v A very powerful feature of SQL: a WHERE clause can itself contain an SQL query! (Actually, so can FROM and HAVING clauses.) v To find sailors who’ve not reserved #103, use NOT IN. v To understand semantics of nested queries, think of a nested loops evaluation: For each Sailors tuple, check the qualification by computing the subquery.

12. Introduction to Database Systems 12 Similar, but Different SELECT * FROM Sailors S WHERE S.sname IN (SELECT Y.sname FROM YoungSailors Y) SELECT S.* FROM Sailors S, YoungSailors Y WHERE S.sname = Y.sname

13. Introduction to Database Systems 13 Nested Queries with Correlation 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: v EXISTS is another set comparison operator, like IN. v If UNIQUE is used, and * is replaced by R.bid, finds sailors with at most one reservation for boat #103. ( UNIQUE checks for duplicate tuples; * denotes all attributes. Why do we have to replace * by R.bid ?)

14. Introduction to Database Systems 14 More on Set-Comparison Operators v We’ve already seen IN, EXISTS and UNIQUE. Can also use NOT IN, NOT EXISTS and NOT UNIQUE. v Also available: op ANY, op ALL, op IN v Find sailors whose rating is greater than that of some sailor called Horatio: SELECT * FROM Sailors S WHERE S.rating > ANY ( SELECT S2.rating FROM Sailors S2 WHERE S2.sname=‘Horatio’)

15. Introduction to Database Systems 15 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’) v Similarly, EXCEPT queries re-written using NOT IN. v To find names (not sid ’s) of Sailors who’ve reserved both red and green boats, just replace S.sid by S.sname in SELECT clause. (What about INTERSECT query?)

16. Introduction to Database Systems 16 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)) v Let’s do it the hard way, without EXCEPT : Sailors S such that... there is no boat B without... a Reserves tuple showing S reserved B 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 who’ve reserved all boats...

17. Introduction to Database Systems 17 Aggregate Operators v 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) single column

18. Introduction to Database Systems 18 Find name and age of the oldest sailor(s) SELECT S.sname, MAX (S.age) FROM Sailors S v The first query is illegal! SELECT S.sname, S.age FROM Sailors S WHERE S.age = ( SELECT MAX (S2.age) FROM Sailors S2)

19. Introduction to Database Systems 19 GROUP BY and HAVING v So far, we’ve applied aggregate operators to all (qualifying) tuples. Sometimes, we want to apply them to each of several groups of tuples. v Consider: Find the age of the youngest sailor for each rating level. –In general, we don’t know how many rating levels exist, and what the rating values for these levels are! –Suppose we know that rating values go from 1 to 10; we can write 10 queries that look like this (!): SELECT MIN (S.age) FROM Sailors S WHERE S.rating = i For i = 1, 2,..., 10:

20. Introduction to Database Systems 20 Queries With GROUP BY and HAVING v The target-list contains (i) attribute names (ii) terms with aggregate operations (e.g., MIN ( S.age )). –The attribute list (i) must be a subset of grouping-list. SELECT [DISTINCT] target-list FROM relation-list WHERE qualification GROUP BY grouping-list HAVING group-qualification ORDER BY subset-of-target-list

21. Introduction to Database Systems 21 Conceptual Evaluation v Cross-product of relation-list is computed; tuples that fail qualification are discarded; ` unnecessary’ fields are deleted; remaining tuples are partitioned into groups by the value of attributes in grouping-list. v The group-qualification is then applied to eliminate some groups. Expressions in group-qualification must have a single value per group ! –In effect, an attribute in group-qualification that is not an argument of an aggregate op also appears in grouping-list. v One answer tuple is generated per qualifying group.

22. Introduction to Database Systems 22 Find the age of the youngest sailor with age 18, for each rating with at least 2 such sailors SELECT S.rating, MIN (S.age) FROM Sailors S WHERE S.age >= 18 GROUP BY S.rating HAVING COUNT (*) > 1 v Only S.rating and S.age are mentioned in the SELECT, GROUP BY or HAVING clauses; Answer relation

23. Introduction to Database Systems 23 For each red boat, find the number of reservations for this boat SELECT B.bid, COUNT (*) AS scount FROM Boats B, Reserves R WHERE R.bid=B.bid AND B.color=‘red’ GROUP BY B.bid v What do we get if we remove B.color=‘red’ from the WHERE clause and add a HAVING clause with this condition?

24. Introduction to Database Systems 24 Find the age of the youngest sailor with age > 18, for each rating with at least 2 sailors (of any age) 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) v Just like the WHERE clause, the HAVING clause can also contain a subquery. v Compare this with the query where we considered only ratings with 2 sailors over 18!

25. Introduction to Database Systems 25 Find those ratings for which the average age is the minimum over all ratings v Aggregate operations cannot be nested! WRONG: SELECT S.rating FROM Sailors S WHERE S.rating = ( 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) v Correct solution (in SQL/92): View or Table Expr.

26. Introduction to Database Systems 26 Creating Views v A view is just a relation, but we store a definition, rather than a set of tuples. CREATE VIEW ActiveSailors (name, age, day) AS SELECT S.sname, S.age, R.day FROM Sailors S, Reserves R WHERE S.name=R.sname AND S.rating>6 v Views can be dropped using the DROP VIEW command. v Views implement a conceptual schema (with security and information hiding)

27. Introduction to Database Systems 27 Queries on Views v Evaluated using a technique known as query modification v Note how sname has been renamed to name to match the view definition. SELECT A.name, MAX ( A.day ) FROM Active Sailors A GROUP BY A.name SELECT name, MAX ( A.Day ) FROM ( SELECT S.sname AS name, S.age, R.day FROM Sailors S, Reserves R WHERE S.sname=R.sname AND S.rating>6 ) AS A GROUP BY A.name

28. Introduction to Database Systems 28 Creating and Deleting Relations CREATE TABLE Boats (bid: INTEGER, bname: CHAR(10), color: CHAR(10) ) CREATE TABLE Reserves (sname: CHAR(10), bid: INTEGER, day: DATE) DROP TABLE Boats ALTER TABLE Boats ADD COLUMN boatkind: CHAR(10)

29. Introduction to Database Systems 29 Inserting New Records v Single record insertion: INSERT INTO Sailors (sid, sname, rating, age) VALUES (12, ‘Emmanuel’, 5, 21.0) v Multiple record insertion: INSERT INTO Sailors (sid, sname, rating, age) SELECT S.sid, S.name, null, S.age FROM Students S WHERE S.age >= 18

30. Introduction to Database Systems 30 Deleting Records v Can delete all tuples that satisfy condition in a WHERE clause: DELETE FROM Sailors S WHERE S.rating = 5 v Example deletes all unrated sailors; WHERE clause can contain nested queries etc., in general.

31. Introduction to Database Systems 31 Modifying Records v UPDATE command used to modify fields of existing tuples. v WHERE clause is applied first and determines fields to be modified. SET clause determines new values. v If field being modified is also used to determine new value, value on rhs is old value. UPDATE Sailors S SET S.rating=S.rating-1 WHERE S.age < 15 UPDATE Sailors S SET S.rating=S.rating-1 WHERE S.rating >= 8

32. Introduction to Database Systems 32 Primary and Candidate Keys CREATE TABLE Reserves ( sname CHAR (10) bid INTEGER, day DATE, PRIMARY KEY (sname, bid, day) ) CREATE TABLE Reserves ( sname CHAR (10) NOT NULL, bid INTEGER, day DATE, PRIMARY KEY (bid, day) UNIQUE (sname) )

33. Introduction to Database Systems 33 Foreign Keys v Primary key fields cannot contain null values. v Foreign key field values must match values in some S tuple, or be null. CREATE TABLE Boats ( bid INTEGER, bname CHAR (10) color CHAR(10), PRIMARY KEY (bid) ) CREATE TABLE Reserves ( sname CHAR (10) NOT NULL, bid INTEGER, day DATE, PRIMARY KEY (bid, day) FOREIGN KEY (bid) REFERENCES Boats )

34. Introduction to Database Systems 34 Enforcing Referential Integrity v What should be done if a Reserves tuple with a non- existent boat id is inserted? ( Reject it! ) v What should be done if a Boats tuple is deleted? –Disallow the deletion. –Also delete all Reserves tuples that refer to it. –Set bid of Reserves tuples that refer to it to a default bid. –Set bid of Reserves tuples that refer to it to null. v What if the primary key of a Boats tuple is updated?

35. Introduction to Database Systems 35 Referential Integrity in SQL/92 v CREATE TABLE Reserves v ( sname CHAR (10) NOT NULL, v bid INTEGER DEFAULT 1000, v day DATE, v PRIMARY KEY (bid, day) v UNIQUE (sname) v FOREIGN KEY (bid) v REFERENCES Boats v ON DELETE CASCADE v ON UPDATE SET DEFAULT )

36. Introduction to Database Systems 36 Updates on Views R S –View update  updating the underlying relations. –Sometimes ambiguous or even impossible! –E.g.: delete (just) the highlighted tuple from instance A of view ActiveSailors. A

37. Introduction to Database Systems 37 Embedded SQL: An Example char SQLSTATE[6]; EXEC SQL BEGIN DECLARE SECTION char c_sname[20]; short c_minrating; float c_age; EXEC SQL END DECLARE SECTION c_minrating = random(); EXEC SQL DECLARE sinfo CURSOR FOR SELECT S.sname, S.ageFROM Sailors S WHERE S.rating > :c_minrating; do { EXEC SQL FETCH sinfo INTO :c_sname, :c_age; printf(“%s is %d years old\n”, c_sname, c_age); } while (SQLSTATE != ‘02000’); EXEC SQL CLOSE sinfo;

38. Introduction to Database Systems 38 Null Values v 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. v The presence of null complicates many issues. E.g.: –Special predicates needed to check if value is/is not null. –How to represent nulls on disk, in memory? –Is rating>8 true or false when rating is equal to null ? What about AND, OR and NOT connectives? –We need a 3-valued logic (true, false and unknown ). –WHERE clause eliminates rows that don’t evaluate to true.