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Relational Algebra and SQL Exercises Dr. Shiyong Lu Department of Computer Science Wayne State University Shiyong@wayne.edu ©copyright 2007, all rights reserved

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Database schema for the exercises Professor(ssn, profname, status, salary) Course(crscode, crsname, credits) Taught(crscode, semester, ssn) Assumption: (1) Each course has only one instructor in each semester; (2) all professors have different salaries; (3) all professors have different names; (4) all courses have different names; (5) status can take values from “Full”, “Associate”, and “Assistant”.

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Query 1 Return those professors who have taught ‘csc6710’ but never ‘csc7710’.

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Relational Algebra Solution ssn ( crscode=‘csc6710’ (Taught))- ssn ( crscode=‘csc7710’ (Taught))

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SQL Solution (SELECT ssn From Taught Where crscode = ‘CSC6710’) EXCEPT (SELECT ssn From Taught Where crscode = ‘CSC7710’))

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Query 2 Return those professors who have taught both ‘csc6710’ and ‘csc7710’.

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Relational Algebra Solution ssn ( crscode=‘csc6710’ crscode=‘csc7710’ (Taught), wrong! ssn ( crscode=‘csc6710’ (Taught)) ssn ( crscode=‘csc7710’ (Taught)), correct!

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SQL Solution SELECT T1.ssn From Taught T1, Taught T2, Where T1.crscode = ‘CSC6710’ AND T2.crscode=‘CSC7710’ AND T1.ssn=T2.ssn

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Query 3 Return those professors who have never taught ‘csc7710’.

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Relational Algebra Solution ssn ( crscode<>‘csc7710’ (Taught)), wrong answer! ssn (Professor)- ssn ( crscode=‘csc7710’ (Taught)), correct answer!

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SQL Solution (SELECT ssn From Professor) EXCEPT (SELECT ssn From Taught T Where T.crscode = ‘CSC7710’)

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Query 4 Return those professors who taught ‘CSC6710’ and ‘CSC7710” in the same semester

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Relational Algebra Solution ssn ( crscode1=‘csc6710’ (Taught[crscode1, ssn, semester]) crscode2=‘csc7710’ (Taught[crscode2, ssn, semester])) Relational Algebra Solution

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SQL Solution SELECT T1.ssn From Taught T1, Taught T2, Where T1.crscode = ‘CSC6710’ AND T2.crscode=‘CSC7710’ AND T1.ssn=T2.ssn AND T1.semester=T2.semester

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Query 5 Return those professors who taught ‘CSC6710’ or ‘CSC7710” but not both.

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Relational Algebra Solution ssn ( crscode=‘csc6710’ crscode=‘csc7710’ (Taught))- ( ssn ( crscode=‘csc6710’ (Taught)) ssn ( crscode=‘csc7710’ (Taught)))

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SQL Solution (SELECT ssn FROM Taught T WHERE T.crscode=‘CSC6710’ OR T.crscode=‘CSC7710’) Except (SELECT T1.ssn From Taught T1, Taught T2, Where T1.crscode = ‘CSC6710’) AND T2.crscode=‘CSC7710’ AND T1.ssn=T2.ssn)

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Query 6 Return those courses that have never been taught.

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Relational Algebra Solution crscode (Course)- crscode (Taught)

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SQL Solution (SELECT crscode FROM Course) EXCEPT (SELECT crscode FROM TAUGHT )

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Query 7 Return those courses that have been taught at least in two semesters.

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Relational Algebra Solution crscode ( semester1 <> semester2 ( Taught[crscode, ssn1, semester1] Taught[crscode, ssn2, semester2]))

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SQL Solution SELECT T1.crscode FROM Taught T1, Taught T2 WHERE T1.crscode=T2.crscode AND T1.semester <> T2.semester

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Query 8 Return those courses that have been taught at least in 10 semesters.

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SQL Solution SELECT crscode FROM Taught GROUP BY crscode HAVING COUNT(*) >= 10

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Query 9 Return those courses that have been taught by at least 5 different professors.

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SQL Solution SELECT crscode FROM (SELECT DISTINCT crscode, ssn FROM TAUGHT) GROUP BY crscode HAVING COUNT(*) >= 5 SELECT crscode FROM Course C WHERE (SELECT COUNT(DISTINCT *) FROM Taught T WHERE T.crscode = C.crscode ) >=5.

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Query 10 Return the names of professors who ever taught ‘CSC6710’.

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Relational Algebra Solution profname ( crscode=‘csc6710’ (Taught) Professor)

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SQL Solution SELECT P.profname FROM Professor P, Taught T WHERE P.ssn = T.ssn AND T.crscode = ‘CSC6710’

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Query 11 Return the names of full professors who ever taught ‘CSC6710’.

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Relational Algebra Solution profname ( crscode=‘csc6710’ (Taught) status=‘full’ (Professor))

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SQL Solution SELECT P.profname FROM Professor P, Taught T WHERE P.status = ‘full’ AND P.ssn = T.ssn AND T.crscode = ‘CSC6710’

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Query 12 Return the names of full professors who ever taught at least two courses in one semester.

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Relational Algebra Solution profname ( ssn ( crscode1 <> crscode2 ( Taught[crscode1, ssn, semester] Taught[crscode2, ssn, semester]))) status=‘full’ (Professor))

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SQL Solution SELECT P.profname FROM Professor P, Taught T1, Taught T2 WHERE P.status = ‘Full’ AND P.ssn = T1.ssn AND T1.ssn = T2.ssn AND T1.crscode <> T2.crscode AND T1.semester = T2.semester

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SQL Solution SELECT P.profname FROM Professor P WHERE status = ‘Full’ AND ssn IN( SELECT ssn FROM Taught GROUP BY ssn, semester HAVING COUNT(*) >= 2 )

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Query 13 Delete those professors who never taught a course.

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SQL Solution DELETE FROM Professor WHERE ssn NOT IN (SELECT ssn FROM Taught )

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SQL Solution DELETE FROM Professor WHERE ssn IN ( (SELECT ssn FROM Professor) EXCEPT (SELECT ssn FROM Taught) )

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SQL Solution DELETE FROM Professor P WHERE NOT EXISTS ( SELECT * FROM Taught T WHERE T.ssn = P.ssn )

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Query 14 Change all the credits to 4 for those courses that are taught in f2006 semester.

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SQL Solution UPDATE Course SET credits = 4 WHERE crscode IN ( SELECT crscode FROM Taught WHERE semester = ‘f2006’ )

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Query 15 Return the names of the professors who have taught more than 30 credits of courses.

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SQL Solution SELECT profname FROM Professor WHERE ssn IN ( SELECT T.ssn FROM Taught T, Course C WHERE T.crscode = C.crscode GROUP BY T.ssn HAVING SUM(C.credits) > 30 )

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Query 16 Return the name(s) of the professor(s) who taught the most number of courses in S2006.

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SQL Solution SELECT profname FROM Professor WHERE ssn IN( SELECT ssn FROM Taught WHERE semester = ‘S2006’ GROUP BY ssn HAVING COUNT(*) = (SELECT MAX(Num) FROM (SELECT ssn, COUNT(*) as Num FROM Taught WHERE semester = ‘S2006’ GROUP BY ssn) )

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Query 17 List all the course names that professor ‘Smith” taught in Fall of 2007.

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Relational Algebra Solution crsname ( profname=‘Smith’ (Professor) semester=‘f2007’ (Taught) Course)

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SQL Solution SELECT crsname FROM Professor P, Taught T, Course C WHERE P.profname = ‘Smith’ AND P.ssn = T.ssn AND T.semester = ‘F2007’ AND T.crscode = C.crscode

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Query 18 In chronological order, list the number of courses that the professor with ssn ssn = 123456789 taught in each semester.

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SQL Solution SELECT semester, COUNT(*) FROM Taught WHERE ssn = ‘123456789’ GROUP BY semester ORDER BY semester ASC

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Query 19 In alphabetical order of the names of professors, list the name of each professor and the total number of courses she/he has taught.

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SQL Solution SELECT P.profname, COUNT(*) FROM Professor P, Taught T WHERE P.ssn = T.ssn GROUP BY P.ssn, P.profname ORDER BY P.profname ASC

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Query 20 Delete those professors who taught less than 10 courses.

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SQL Solution DELETE FROM Professor WHERE ssn IN( SELECT ssn FROM Taught GROUP BY ssn HAVING COUNT(*) < 10 )

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Query 21 Delete those professors who taught less than 40 credits.

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SQL Solution DELETE FROM Professor WHERE ssn IN( SELECT T.ssn FROM Taught T, Course C WHERE T.crscode = C.crscode GROUP BY ssn HAVING SUM(C.credits) < 40 )

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Query 22 List those professors who have not taught any course in the past three semesters (F2006, W2007, F2007).

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SQL Solution SELECT * FROM Professor P WHERE NOT EXISTS( SELECT * FROM Taught WHERE P.ssn = T.ssn AND (T.semester = ‘F2006’ OR T.semester = ‘W2007’ OR T.semester=‘F2007’)) )

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Query 23 List the names of those courses that professor Smith have never taught.

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Relational Algebra Solution crsname (Course)- crsname ( profname=‘Smith’ (Professor) (Taught) Course)

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SQL Solution SELECT crsname FROM Course C WHERE NOT EXISTS SELECT * FROM Professor P, Taught T WHERE P.profname=‘Smith’ AND P.ssn = T.ssn AND T.crscode = C.crscode )

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Query 24 Return those courses that have been taught by all professors.

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Relational Algebra Solution crscode, ssn (Taught)/ ssn (Professor)

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SQL Solution SELECT crscode FROM Taught T1 WHERE NOT EXISTS( (SELECT ssn FROM Professor) EXCEPT (SELECT ssn FROM Taught T2 WHERE T2.crscode = T1.crscode) )

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Query 25 Return those courses that have been taught in all semesters.

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Relational Algebra Solution crscode, semester (Taught)/ semester (Taught)

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SQL Solution SELECT crscode FROM Taught T1 WHERE NOT EXISTS( (SELECT semester FROM Taught) EXCEPT (SELECT semester FROM Taught T2 WHERE T2.crscode = T1.crscode) )

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Query 26 Return those courses that have been taught ONLY by assisitant professors.

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Relational Algebra Solution crscode (Course) - crscode ( status ‘Assistant’ (Professor) Taught)

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SQL Solution SELECT crscode FROM Course C WHERE c.crscode NOT IN( (SELECT crscode FROM Taught T, Professor P WHERE T.ssn = P.ssn AND P.status=‘Junior’ )

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Query 27 Return the professors who taught the largest number of courses in Fall 2001.

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SQL Solution SELECT * FROM Professor P1 WHERE Not EXISTS ( SELECT * FROM Professor P2 WHERE( (SELECT COUNT(*) FROM Taught WHERE Taught.ssn = P2.ssn AND Taught.semester=‘F2001’) > (SELECT COUNT(*) FROM Taught WHERE Taught.ssn = P1.ssn AND Taught.semester=‘F2001’) )

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Query 28 Return the professor who earns the highest salary.

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SQL Solution SELECT * FROM Professor WHERE salary = ( (SELECT MAX(salary) FROM Professor P )

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Query 29 Return the professor who earns the second highest salary.

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SQL Solution SELECT * FROM Professor P1 WHERE 1 = ( (SELECT COUNT(*) FROM Professor P2 WHERE P2.salary > P1.salary ) Problem: return the professor who earns the 10 th highest salary.

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