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Schedule

Schedule. Today: Subqueries, Grouping and Aggregation. Read Sections 6.3-6.4. Next Modifications, Schemas, Views. Read Sections 6.5-6.7. After that Constraints. Read Sections 7.1-7.3, 7.4.1. Union, Intersection, Difference.

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Schedule

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  1. Schedule • Today: • Subqueries, Grouping and Aggregation. • Read Sections 6.3-6.4. • Next • Modifications, Schemas, Views. • Read Sections 6.5-6.7. • After that • Constraints. • Read Sections 7.1-7.3, 7.4.1. Holliday - coen 178

  2. Union, Intersection, Difference “(subquery) UNION (subquery)” produces the union of the two relations. • Similarly for INTERSECT, and EXCEPT = set difference. • But: in Oracle set difference is MINUS, not EXCEPT. Example Find the drinkers and beers such that the drinker likes the beer and frequents a bar that serves it. Likes(drinker, beer) Sells(bar, beer, price) Frequents(drinker, bar) (SELECT * FROM Likes) INTERSECT (SELECT drinker, beer FROM Sells, Frequents WHERE Frequents.bar = Sells.bar); Holliday - coen 178

  3. Forcing Set Semantics • Default for select-from-where is bag; default for union, intersection, and difference is set. • Force set semantics with DISTINCT after SELECT. Find the different prices charged for beers. Sells(bar, beer, price) SELECT DISTINCT price FROM Sells; Holliday - coen 178

  4. Forcing Bag Semantics • Force bag semantics with ALL • In the case of set operations, duplicates are automatically eliminated. To retain duplicates, useunion all, intersect all, and except all. Holliday - coen 178

  5. More on Bags • There may be duplicate tuples in the results of SQL queries. To determine the number of duplicate tuples in the result. 1. If there are c1 copies of a tuple t1 in r and t1 satisfies the predicate P, then there will be c1 copies of t1 in select * from r where P • If there are c1 copies of a tuple t1 in r, then there will be c1 copies of select A from t1 in select A from r • If there are c1 copies of a tuple t1 in r and c2 copies of t2 in s, then there will be c1* c2 copies of the tuple t1.t2 in select * from r, s Holliday - coen 178

  6. How many tuples? select * r= from s where C = 3 select B from r select B s= from r, s select B select A from r, s from r,s where C = 3 where A=C Holliday - coen 178

  7. Forcing Set Semantics with DISTINCT select distinct B r= from r select distinct B from r, s s= select distinct C from s Holliday - coen 178

  8. A B 1 a 2 a C 2 3 3 How many tuples? (C=3)(s)r= B(r) B(rs)s= B (C=3)(rs) A (C=3)(rs) Holliday - coen 178

  9. Exercise: Use set operations to- • Find all customers who have both a loan and an account. • Find all customers who have an account but no loan. Depositor = (customer-name, account#) Borrower = (customer-name, loan#) Holliday - coen 178

  10. Answers (select customer-name from Depositor) intersect (select customer-name from Borrower) (select customer-name from Depositor) except (select customer-name from Borrower) Holliday - coen 178

  11. Join-Based Expressions A number of forms are provided. • Can be used either stand-alone (in place of a select-from-where) or to define a relation in the FROM-clause. R NATURAL JOIN S (delete duplicate attribute) R JOIN S ONcondition (cross product + ) e.g., condition:R.B=S.B R CROSS JOIN S (cross product) R OUTER JOIN S • Outerjoin can be modified by: 1. Optional NATURAL in front. 2. Optional ON condition at end. 3. Optional LEFT, RIGHT, or FULL (default) before OUTER. • LEFT = pad (with NULL) dangling tuples of R only; RIGHT = pad dangling tuples of S only. Holliday - coen 178

  12. D A B B C Smith a 101 101 35 b Jones 103 102 24 Johnson b 104 103 65 Join Exercise R = S = What is R natural join S? (no second B) Holliday - coen 178

  13. D A B B C Smith a 101 101 35 b Jones 103 102 24 Johnson b 104 103 65 Outer Join Exercise R = S = What is R natural left outer join S? Holliday - coen 178

  14. D A B B C Smith a 101 101 35 b Jones 103 102 24 Johnson b 104 103 65 Outer Join Exercise R = S = What is R natural right outer join S? Holliday - coen 178

  15. D A B B C Smith a 101 101 35 b Jones 103 102 24 Johnson b 104 103 65 Full Outer Join Exercise R = S = What is R natural full outer join S? Holliday - coen 178

  16. D A B B C Smith a 101 101 35 b Jones 103 102 24 Johnson b 104 103 65 Outer Join Exercise R = S = R left outer join S on R.B=S.B Holliday - coen 178

  17. Aggregations Sum, avg, min, max, and count apply to attributes (columns). Also, count(*) applies to tuples. Use these in lists following SELECT. (also having) Output has only one tuple Sells(bar,beer,price) Find the average price of Bud. SELECT AVG(price) FROM Sells WHERE beer = 'Bud'; Holliday - coen 178

  18. Eliminating DuplicatesBefore Aggregation Find the number of different prices at which Bud is sold. Sells(bar, beer, price) SELECT COUNT(DISTINCT price) FROM Sells WHERE beer = 'Bud'; • DISTINCT may be used in any aggregation, but typically only makes sense with COUNT. Holliday - coen 178

  19. Example Using AVG • Find the average account balance at the Oakland branch. Account = (branch-name,account#,balance) select avg(balance) from Account where branch-name = "Oakland“ What does the result look like? Holliday - coen 178

  20. Example Using AVG • Find the average account balance at the Oakland branch of accounts over $3000. • Account = (branch-name, account#, balance) select avg(balance) from Account where branch-name = "Oakland“ and balance>3000 Holliday - coen 178

  21. Grouping Follow select-from-where by GROUP BY and a list of attributes. • The relation that is the result of the FROM and WHERE clauses is grouped according to the values of these attributes, and aggregations take place only within a group. Example:Find the average sales price for each beer. Sells(bar, beer, price) SELECT beer, AVG(price) FROM Sells GROUP BY beer; Holliday - coen 178

  22. Avg sales price for each beer SELECT beer, AVG(price) FROM Sells GROUP BY beer; Holliday - coen 178

  23. Example Find, for each drinker, the average price of Bud at the bars they frequent. Sells(bar, beer, price) Frequents(drinker, bar) SELECT drinker, AVG(price) FROM Frequents, Sells WHERE beer = 'Bud' AND Frequents.bar = Sells.bar GROUP BY drinker; Holliday - coen 178

  24. When Grouping… • Note: grouping occurs after the cross product in FROM and the condition in the WHERE is performed. • When rows (tuples) are grouped, one line of output is produced for each group. Holliday - coen 178

  25. Restriction on SELECT Lists With Aggregation • If any aggregation is used, then each element of a SELECT clause must either be aggregated or appear in a group-by clause (if a where clause is present). • The following might seem a tempting way to find the bar that sells Bud the cheapest: Sells(bar, beer, price) SELECT bar, MIN(price) FROM Sells WHERE beer = 'Bud'; • But it is illegal in most SQL implementations. Holliday - coen 178

  26. HAVING Clauses HAVING clauses are selections on groups, just as WHERE clauses are selections on tuples. • Condition can use the tuple variables or relations in the FROM and their attributes, just like the WHERE can. • But the tuple variables range only over the group. • And the attribute better make sense within a group; i.e., be one of the grouping attributes. Holliday - coen 178

  27. Example Find the average price of those beers that are either served in at least 3 bars or manufactured by Anheuser-Busch. Beers(name, manf) Sells(bar, beer, price) SELECT beer, AVG(price) FROM Sells GROUP BY beer HAVING COUNT(*) >= 3 OR beer IN ( SELECT name FROM Beers WHERE manf = 'Anheuser-Busch' ); Holliday - coen 178

  28. Before we go on.. • How do we find the bar that sells Bud the cheapest? Sells(bar, beer, price) Select bar From Sells Where beer = ‘Bud’ and price = (select min(price) from Sells where beer = ‘Bud’ ) Holliday - coen 178

  29. Alternatively SELECT bar FROM Sells WHERE beer = ‘Bud’ and price <= ALL( SELECT price FROM Sells WHERE beer = ‘Bud’); Holliday - coen 178

  30. Database Modifications • So far, we have looked at queries that ask about the current state of the database (instance). • We use similar syntax to make changes to the database. • Modification = insert + delete + update. Holliday - coen 178

  31. DB Insert Insertion of a Tuple INSERT INTO relation VALUES (list of values). • Inserts the tuple = list of values, associating values with attributes in the order the attributes were declared. • Forget the order? List the attributes as arguments of the relation. Example Likes(drinker, beer) Insert the fact that Sally likes Bud. INSERT INTO Likes(drinker, beer) VALUES('Sally', 'Bud'); Holliday - coen 178

  32. Insertion of the Result of a Query INSERT INTO relation (subquery). Example Create a (unary) table of all Sally's buddies, i.e., the people who frequent bars that Sally also frequents. Frequents(drinker, bar) CREATE TABLE Buddies( name char(30) ); INSERT INTO Buddies (SELECT DISTINCT d2.drinker FROM Frequents d1, Frequents d2 WHERE d1.drinker = 'Sally' AND d2.drinker <> 'Sally' AND d1.bar = d2.bar ); Holliday - coen 178

  33. Deletion DELETE FROM relation WHERE condition. • Deletes all tuples satisfying the condition from the named relation. • Sally no longer likes Bud. Likes(drinker, beer) DELETE FROM Likes WHERE drinker = 'Sally' AND beer = 'Bud'; • Make the Likes relation empty. DELETE FROM Likes; Holliday - coen 178

  34. Example • Delete all beers for which there is another beer by the same manufacturer. Beers(name,manf) DELETE FROM Beers b WHERE EXISTS (SELECT name FROM Beers WHERE manf = b.manf AND name <> b.name ); • Note alias for relation from which deletion occurs. • Subquery evaluated once for each row of b Holliday - coen 178

  35. Semantics is tricky. If A.B. makes Bud and BudLite (only), does deletion of Bud make BudLite not satisfy the condition? • SQL semantics: all conditions in modifications must be evaluated by the system before any changes due to that command occur. • In Bud/Budlite example, we would first identify both beers as targets, and then delete both. Holliday - coen 178

  36. More on Delete • Oracle 8i does not allow complex conditions in the where clause. • You can only delete from one table at a time. Delete all accounts at every branch located in Fremont.  delete from Account where branch-name in (select branch-name from Branch where branch-city="Fremont" ) Holliday - coen 178

  37. Updates UPDATE relation SET list of assignments WHEREcondition Example Drinker Fred's phone number is now 555-1212. Drinkers(name, addr, phone) UPDATE Drinkers SET phone = '555-1212' WHERE name = 'Fred'; Holliday - coen 178

  38. Example - Update Make $4 the maximum price for beer. Updates many tuples at once. Sells(bar, beer, price) UPDATE Sells SET price = 4.00 WHERE price > 4.00; Holliday - coen 178

  39. Review/Quiz select branch-name, avg(balance) from Account where account# > 5000 group by branch-name havingavg(balance) > 2000 Holliday - coen 178

  40. Answer • If both a where clause and a having command are present, the where clause is done first. Holliday - coen 178

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