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COURSE -----JAV OAU SQL SQL SQL>
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BEGINDATE COUNT(*) ----------- -------13-DEC-1999 5 10-AUG-1999 3 12-APR-1999 4 04-OCT-1999 3
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APPENDIX D ANSWERS TO THE EXERCISES
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In this case, accessing the REGISTRATIONS table is enough, because you are not interested in offerings without registrations. The solution would have been more complicated if the question were ... with fewer than three registrations, because zero is also less than three. 6. Provide the employee numbers of all employees who ever taught a course as a trainer, but never attended a course as an attendee. Solution 8-6a. First Solution SQL> select trainer from offerings 2 minus 3 select attendee from registrations; TRAINER -------7369
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SQL> This solution looks good; however, if you look very carefully, the solution is suspect. You don t see it immediately, but this result doesn t contain a single row, but two rows, as becomes apparent if you set FEEDBACK to 1: SQL> set feedback 1 SQL> / TRAINER -------7369
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2 rows selected. SQL> Because a null value obviously doesn t represent a valid trainer, you need to exclude null values in the TRAINER column explicitly. Solution 8-6b. Second Solution, Excluding Null Values SQL> 2 3 4 select trainer from offerings where trainer is not null minus select attendee from registrations;
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APPENDIX D ANSWERS TO THE EXERCISES
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TRAINER -------7369 1 row selected. SQL> 7. Which employees attended a specific course more than once Solution 8-7. SQL> 2 3 4 select from group by having attendee,course registrations attendee,course count(*) > 1 ;
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ATTENDEE -------7698 7788 7902 SQL>
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COURSE -----SQL JAV SQL
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8. For all trainers, provide their name and initials, the number of courses they taught, the total number of students they had in their classes, and the average evaluation rating. Round the evaluation ratings to one decimal. Solution 8-8. SQL> 2 3 4 5 6 7 8 9 10 11 t.init, t.ename count(distinct begindate) courses count(*) attendees round(avg(evaluation),1) evaluation employees t registrations r join offerings o using (course, begindate) where t.empno = o.trainer group by t.init, t.ename; select , , , from ,
APPENDIX D ANSWERS TO THE EXERCISES
INIT ----N AA JM MG SCJ SQL>
ENAME COURSES ATTENDEES EVALUATION -------- -------- --------- ---------SMITH 3 7 4 ADAMS 1 3 4 JONES 2 8 4.3 FORD 2 5 4 SCOTT 1 3
Note While counting courses, this solution assumes that trainers cannot teach more than one
course on the same day.
9. List the name and initials of all trainers who ever had their own manager as a student in a general course (category GEN). Solution 8-9. SQL> 2 3 4 5 6 7 8 9 10 11 select from , , , where and and and and and distinct e.ename, e.init employees e courses c offerings o registrations r e.empno = o.trainer e.mgr = r.attendee c.code = o.course o.course = r.course o.begindate = r.begindate c.category = 'GEN';
ENAME INIT -------- ----SMITH N SQL> 10. Did we ever use two classrooms at the same time in the same course location Solution 8-10. SQL> select o1.location 2 , o1.begindate, o1.course, c1.duration 3 , o2.begindate, o2.course
APPENDIX D ANSWERS TO THE EXERCISES
4 5 6 7 8 9 10 11 12
from offerings o1 , offerings o2 , courses c1 where o1.location = o2.location and (o1.begindate < o2.begindate or o1.course <> o2.course ) and o1.course = c1.code and o2.begindate between o1.begindate and o1.begindate + c1.duration;
LOCATION BEGINDATE COUR DURATION BEGINDATE COURSE -------- ----------- ---- -------- ----------- -----DALLAS 01-FEB-2000 JAV 4 03-FEB-2000 XML SQL> The solution searches for two different course offerings (see lines 8 and 9) at the same location (see line 7) overlapping each other (see lines 11 and 12). Apparently, the Java course starting February 1, 2000, in Dallas overlaps with the XML course starting two days later (note that the Java course takes four days). 11. Produce a matrix report (one column per department, one row for each job) where each cell shows the number of employees for a specific department and a specific job. In a single SQL statement, it is impossible to dynamically derive the number of columns needed, so you may assume you have three departments only: 10, 20, and 30. Solution 8-11. SQL> 2 3 4 5 6 7 8 9 10 11 12 13 14 job count(case when deptno <> 10 then null else deptno end ) as dept_10 , sum(case deptno when 20 then 1 else 0 end ) as dept_20 , sum(decode(deptno,30,1,0)) as dept_30 from employees group by job; select ,
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