数据库系统基础教程第八章答案.doc
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Section 1 Exercise 8.1.1 a) CREATE VIEW RichExec AS SELECT * FROM MovieExec WHERE netWorth >= 10000000; b) CREATE VIEW StudioPres (name, address, cert#) AS SELECT MovieExec.name, MovieExec.address, MovieExec.cert# FROM MovieExec, Studio WHERE MovieExec.cert# = Studio.presC#; c) CREATE VIEW ExecutiveStar (name, address, gender, birthdate, cert#, netWorth) AS SELECT star.name, star.address, star.gender, star.birthdate, exec.cert#, Worth FROM MovieStar star, MovieExec exec WHERE star.name = exec.name AND star.address = exec.address; Exercise 8.1.2 a) SELECT name from ExecutiveStar WHERE gender = ‘f’; b) SELECT RichExec.name from RichExec, StudioPres where RichExec.name = StudioPres.name; c) SELECT ExecutiveStar.name from ExecutiveStar, StudioPres WHERE ExecutiveSWorth >= 50000000 AND StudioPres.cert# = RichExec.cert#; Section 2 Exercise 8.2.1 The views RichExec and StudioPres are updatable; however, the StudioPres view needs to be created with a subquery. CREATE VIEW StudioPres (name, address, cert#) AS SELECT MovieExec.name, MovieExec.address, MovieExec.cert# FROM MovieExec WHERE MovieExec.cert# IN (SELECT presCt# from Studio); Exercise 8.2.2 a) Yes, the view is updatable. b) CREATE TRIGGER DisneyComedyInsert INSTEAD OF INSERT ON DisneyComedies REFERENCING NEW ROW AS NewRow FOR EACH ROW INSERT INTO Movies(title, year, length, studioName, genre) VALUES(NewRow.title, NewRow.year, NewYear.length, ‘Disney’, ‘comedy’); c) CREATE TRIGGER DisneyComedyUpdate INSTEAD OF UPDATE ON DisneyComedies REFERENCING NEW ROW AS NewRow FOR EACH ROW UPDATE Movies SET length NewRow.length WHERE title = NewRow.title AND year = NEWROW.year AND studionName = ‘Disney’ AND genre = ‘comedy’; Exercise 8.2.3 a) No, the view is not updatable since it is constructed from two different relations. b) CREATE TRIGGER NewPCInsert INSTEAD OF INSERT ON NewPC REFERENCING NEW ROW AS NewRow FOR EACH ROW (INSERT INTO Product VALUES(NewRow.maker, NewRow.model, ‘pc’)) (INSERT INTO PC VALUES(NewRow.model, NewRow.speed, NewRow.ram, NewRow.hd, NewRow.price)); c) CREATE TRIGGER NewPCUpdate INSTEAD OF UPDATE ON NewPC REFERENCING NEW ROW AS NewRow FOR EACH ROW UPDATE PC SET price = NewPC.price where model = NewPC.model; d) CREATE TRIGGER NewPCDelete INSTEAD OF DELETE ON NeePC REFERENCING OLD ROW AS OldRow FOR EACH ROW (DELETE FROM Product WHERE model = OldRow.model) (DELETE FROM PC where model = OldRow.model); Section 3 Exercise 8.3.1 a) CREATE INDEX NameIndex on Studio(name); b) CREATE INDEX AddressIndex on MovieExec(address); c) CREATE INDEX GenreIndex on Movies(genre, length); Section 4 Exercise 8.4.1 Action No Index Star Index Movie Index Both Indexes Q1 100 4 100 4 Q2 100 100 4 4 I 2 4 4 6 Average 2 + 98p1 + 98p2 4 + 96 p2 4 + 96 p1 6 – 2 p1 – 2 p2 Exercise 8.4.2 Q1 = SELECT * FROM Ships WHERE name = n; Q2 = SELECT * FROM Ships WHERE class = c; Q3 = SELECT * FROM Ships WHERE launched = y; I = Inserts Indexes Actions None Name Class Launched Name & Class Name & Launched Class & Launched Three Indexes Q1 50 2 50 50 2 2 50 2 Q2 1 1 2 1 2 1 2 2 Q3 50 50 50 26 50 26 26 26 I 2 4 4 4 6 6 6 8 Average 2 + 48p1 -p2 + 48p3 4 + 46 p3 - 2 p1 - 3 p2 4 + 46p1 - 2p2 + 46p3 4 + 46p1 - 3p2 + 22p3 6 - 4p1 - 4p2 + 44p3 6 - 4p1 - 5p2 + 20p3 6 - 44p1 - 4p2 + 20p3 8 - 6p1 - 6p2 + 18p3 The best choice of indexes (name and launched) has an average cost of 6 - 4p1 - 5p2 + 20p3 per operation. Section 5 Exercise 8.5.1 Updates to movies that involves title or year UPDATE MovieProd SET title = ‘newTitle’ where title=’oldTitle’ AND year = oldYear; UPDATE MovieProd SET year = newYear where title=’oldYitle’ AND year = oldYear; Update to MovieExec involving cert# DELETE FROM MovieProd WHERE (title, year) IN ( SELECT title, year FROM Movies, MovieExec WHERE cert# = oldCert# AND cert# = producerC# ); INSERT INTO MovieProd SELECT title, year, name FROM Movies, MovieExec WHERE cert# = newCert# AND cert# = producerC#; Exercise 8.5.2 Insertions, deletions, and updates to the base tables Product and PC would require a modification of the materialized view. Insertions into Product with type equal to ‘pc’: INSERT INTO NewPC SELECT maker, model, speed, ram, hd, price FROM Product, PC WHERE Product.model = newModel and Product.model = PC.model; Insertions into PC: INSERT INTO NewPC SELECT maker, ‘newModel’, ‘newSpeed’, ‘newRam’, ‘newHd’, ‘newPrice’ FROM Product WHERE model = ‘newModel’; Deletions from Product with type equal to ‘pc’: DELETE FROM NewPC WHERE maker = ‘deletedMaker’ AND model=’deletedModel’; Deletions from PC: DELETE FROM NewPC WHERE model = ‘deletedModel’; Updates to PC: Update NewPC SET speed=PC.speed, ram=PC.ram, hd=PC.hd, price=PC.price FROM PC where model=pc.model; Update to the attribute ‘model’ needs to be treated as a delete and an insert. Updates to Product: Any changes to a Product tuple whose type is ‘pc’ need to be treated as a delete or an insert, or both. Exercise 8.5.3 Modifications to the base tables that would require a modification to the materialized view: inserts and deletes from Ships, deletes from class, updates to a Class’ displacement. Deletions from Ship: UPDATE ShipStats SET displacement=((displacement * count) – (SELECT displacement FROM Classses WHERE class = ‘DeletedShipClass’) ) / (count – 1), count = count – 1 WHERE country = (SELECT country FROM Classes WHERE class=’DeletedShipClass’); Insertions into Ship: Update ShipStat SET displacement=((displacement*count) + (SELECT displacement FROM Classes WHERE class=’InsertedShipClass’) ) / (count + 1), count = count + 1 WHERE country = (SELECT country FROM Classes WHERE classes=’InsertedShipClass); Deletes from Classes: NumRowsDeleted = SELECT count(*) FROM ships WHERE class = ‘DeletedClass’; UPDATE ShipStats SET displacement = (displacement * count) - (DeletedClassDisplacement * NumRowsDeleted)) / (count – NumRowsDeleted), count = count – NumRowsDeleted WHERE country = ‘DeletedClassCountry’; Update to a Class’ displacement: N = SELECT count(*) FROM Ships where class = ‘UpdatedClass’; UPDATE ShipsStat SET displacement = ((displacement * count) + ((oldDisplacement – newDisplacement) * N))/count WHERE country = ‘UpdatedClassCountry’; Exercise 8.5.4 Queries that can be rewritten with the materialized view: Names of stars of movies produced by a certain producer SELECT starName FROM StarsIn, Movies, MovieExec WHERE movieTitle = title AND movieYear = year AND producerC# = cert# AND name = ‘Max Bialystock’; Movies produced by a certain producer SELECT title, year FROM Movies, MovieExec Where producerC# = cert# AND name = ‘George Lucas’; Names of producers that a certain star has worked with SELECT name FROM Movies, MovieExec, StarsIn Where producerC#=cert# AND title=movieTitle AND year=movieYear AND starName=’Carrie Fisher’; The number of movies produced by given producer SELECT count(*) FROM Movies, MovieExec WHERE producerC#=cert# AND name = ‘George Lucas‘; Names of producers who also starred in their own movies SELECT name FROM Movies, StarsIn, MovieExec WHERE producerC#=cert# AND movieTitle = title AND movieYear = year AND MovieExec.name = starName; The number of stars that have starred in movies produced by a certain producer SELECT count(DISTINCT starName) FROM Movies, StarsIn, MovieExec WHERE producerC#=cert# AND movieTitle = title AND movieYear = year AND name ‘George Lucas’; The number of movies produced by each producer SELECT name, count(*) FROM Movies, MovieExec WHERE producerC#=cert# GROUP BY name- 配套讲稿:
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