Database Management Systems, 3rd Edition

Database Management Systems, 3rd Edition

Raghu Ramakrishnan, Johannes Gehrke

Language: English

Pages: 1104

ISBN: 0072465638

Format: PDF / Kindle (mobi) / ePub

Database Management Systems provides comprehensive and up-to-date coverage of the fundamentals of database systems. Coherent explanations and practical examples have made this one of the leading texts in the field. The third edition continues in this tradition, enhancing it with more practical material.

The new edition has been reorganized to allow more flexibility in the way the course is taught. Now, instructors can easily choose whether they would like to teach a course which emphasizes database application development or a course that emphasizes database systems issues. New overview chapters at the beginning of parts make it possible to skip other chapters in the part if you don't want the detail.

More applications and examples have been added throughout the book, including SQL and Oracle examples. The applied flavor is further enhanced by the two new database applications chapters.

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does not ensure that a manager is initially appointed for each department!) This situation is a strong argument The Relational lvfodel 8~ in favor of using the second approach for one-to-many relationships such as Manages, especially when the entity set with the key constraint also has a total participation constraint. Unfortunately, there are many participation constraints that we cannot capture using SQL, short of using table constraints or assertions. Table constraints and assertions can

instances E1, R2, and S3). Then we join with Reserves to identify sid.., of sailors who have reserved OIle of these boats; this gives us sids 22, 31, 64, and 74 over our example instances. Finally, we join (an intermediate relation containing this set of sids) with Sailors to find the names of Sailors with these sids. This gives us the names Dustin, Horatio, and Lubber on the instances E1, R2, and S3. Another equivalent definition is the following: p(Tempboats, (acolor='red'Vcolor='green'

Boats)) 7fsname(Tempboats [><] Reserves [><] Sailors) Let us now consider a very similar query. (Q6) Find the names of sailors who have reserved a red and a green boat. It is tempting to try to do this by simply replacing U by n in the definition of Tempboats: p(Tempboats2, (acolor='red,Eoats) n (O"color='green,Boats)) '7fsname(Tempboats2 [><] Reserves [><] Sailors) However, this solution is incorrect-it instead tries to compute sailors who have reserved a boat that is both red and green. (Since

E Sailors /\3(1, Br, D) E ReseTves /\ 3(Br, BN,'Ted') E Boats)} (Q7) Find the names of sailoT.'! who have TeseTved at least two boat.s. {(N) I 31, T, A( (1, N, T, A) E Sailors /\ ?JBrl, BT2, Dl, D2( (1, Brl, DI) E Reserves /\(1, Br2, D2) E Reserves /\ Brl # Br2))} CHAPTER . fl 124 Note how the repeated use of variable I ensures that the same sailor has reserved both the boats in question. (Q9) Find the names of sailors who have Teserved all boat8. {(N) I ~I, T, A( (I, N, T, A) E Sailors!\

the name of the sailors having that age. However, this query is illegal in SQL-if the SELECT clause uses an aggregate operation, then it must use only aggregate operations unless the query contains a GROUP BY clause! (The intuition behind this restriction should become clear when we discuss the GROUP BY clause in Section 5.5.1.) Therefore, we cannot use MAX (S.age) as well as S.sname in the SELECT clause. We have to use a nested query to compute the desired answer to Q27: SELECT S.sname, S.age

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