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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 • 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
Union, Intersection, Difference “(subquery) UNION (subquery)” produces the union of the two relations. • Similarly for INTERSECT, EXCEPT = intersection and 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
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
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
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
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
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
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
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
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 R JOIN S ON condition (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
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? Holliday - coen 178
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
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
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
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 left outer join S on R.B=S.B? Holliday - coen 178
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 Example Find the average price of Bud.Sells(bar, beer, price) SELECT AVG(price) FROM Sells WHERE beer = 'Bud'; Holliday - coen 178
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
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
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
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
Avg sales price for each beer SELECT beer, AVG(price) FROM Sells GROUP BY beer; Holliday - coen 178
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
When Grouping… • Note: grouping occurs after the cross product in FROM and condition in WHERE is performed. • When rows (tuples) are grouped, one line of output is produced for each group. Holliday - coen 178
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. Example • 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. • What would the result be if it was legal? Holliday - coen 178
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
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
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
Alternatively SELECT bar FROM Sells WHERE beer = ‘Bud’ and price <= ALL( SELECT price FROM Sells WHERE beer = ‘Bud’); Holliday - coen 178
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
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
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
Deletion DELETE FROM relation WHERE condition. • Deletes all tuples satisfying the condition from the named relation. Example Sally no longer likes Bud. Likes(drinker, beer) DELETE FROM Likes WHERE drinker = 'Sally' AND beer = 'Bud'; Example Make the Likes relation empty. DELETE FROM Likes; Holliday - coen 178
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
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 mods due to that mod command occur. • In Bud/Budlite example, we would first identify both beers a targets, and then delete both. Holliday - coen 178
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
Updates UPDATE relation SET list of assignments WHERE condition. Example Drinker Fred's phone number is 555-1212. Drinkers(name, addr, phone) UPDATE Drinkers SET phone = '555-1212' WHERE name = 'Fred'; Holliday - coen 178
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
Review/Quiz select branch-name, avg(balance) from Account where account# > 5000 group by branch-name havingavg(balance) > 2000 Holliday - coen 178
Answer • If both a where clause and a having command are present, the where clause is done first. Holliday - coen 178