Data Mgmt Lab_Part_15

# Data Mgmt Lab_Part_15 - Relational Algebra and Calculus 43...

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Unformatted text preview: Relational Algebra and Calculus 43 ( CE 3 = CE 4 CA 1 = CA 4 CA 2 = CA 4 CA 3 = CA 4)))) } SQL SELECT C1.eid FROM Certified C1, Certified C2, Certified C3 WHERE (C1.eid = C2.eid AND C2.eid = C3.eid AND C1.aid = C2.aid AND C2.aid = C3.aid AND C3.aid = C1.aid) EXCEPT SELECT C4.eid FROM Certified C4, Certified C5, Certified C6, Certified C7, WHERE (C4.eid = C5.eid AND C5.eid = C6.eid AND C6.eid = C7.eid AND C4.aid = C5.aid AND C4.aid = C6.aid AND C4.aid = C7.aid AND C5.aid = C6.aid AND C5.aid = C7.aid AND C6.aid = C7.aid ) This could also be done in SQL using COUNT. 10. This cannot be expressed in relational algebra (or calculus) because there is no operator to sum values. The query can however be expressed in SQL as follows: SELECT SUM (E.salaries) FROM Employees E 11. This cannot be expressed in relational algebra or relational calculus or SQL. The problem is that there is no restriction on the number of intermediate ights. All of the query methods could find if there was a ight directly from Madison to...
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## This note was uploaded on 01/17/2012 for the course EGN 4302 taught by Professor Dr.vishak during the Fall '12 term at University of Central Florida.

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Data Mgmt Lab_Part_15 - Relational Algebra and Calculus 43...

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