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Creating Data Matrix ECC200 in Java ANSWERS TO THE EXERCISES

APPENDIX B
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ANSWERS TO THE EXERCISES
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4 no rows selected SQL>
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offerings o);
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Solution 9-3b. Using NOT EXISTS SQL> select e.* 2 from employees e 3 where not exists (select o.trainer 4 from offerings o 5 where o.trainer = e.empno); EMPNO ----7499 7521 7654 7698 7782 7839 7844 7900 7934 ENAME -------ALLEN WARD MARTIN BLAKE CLARK KING TURNER JONES MILLER INIT ----JAM TF P R AB CC JJ R TJA JOB MGR BDATE MSAL COMM DEPTNO -------- ----- ----------- ----- ----- -----SALESREP 7698 20-FEB-1961 1600 300 30 SALESREP 7698 22-FEB-1962 1250 500 30 SALESREP 7698 28-SEP-1956 1250 1400 30 MANAGER 7839 01-NOV-1963 2850 30 MANAGER 7839 09-JUN-1965 2450 10 DIRECTOR 17-NOV-1952 5000 10 SALESREP 7698 28-SEP-1968 1500 0 30 ADMIN 7698 03-DEC-1969 800 30 ADMIN 7782 23-JAN-1962 1300 10
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9 rows selected. SQL> At first sight, you might think that both of these solutions are correct. However, the results are different. Now, which one is the correct solution You can come up with convincing arguments for both solutions. Note that you have three course offerings with a null value in the TRAINER column. If you interpret these null values as trainer unknown, you can never say with certainty that an employee never taught a course. The second query obviously treats the null values differently. Its result (with nine employees) is what you probably expected.
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The different results are not caused by an SQL bug. You simply have two SQL statements with different results, so they must have a different meaning. In such cases, you must revisit the query in natural language and try to formulate it more precisely in order to eliminate any ambiguities. Last but not least, our OFFERINGS table happens to contain only data from the past. If you want a correct answer to this exercise under all circumstances, you should also add a condition to check the course dates against SYSDATE. 4. Which employees attended all build courses (category BLD) They are entitled to get a discount on the next course they attend.
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APPENDIX B
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ANSWERS TO THE EXERCISES
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Solution 9-4a. Using NOT EXISTS Twice SQL> select 2 from 3 where 4 5 6 7 8 9 10 11 12 13 e.empno, e.ename, e.init employees e not exists (select c.* from courses c where c.category = 'BLD' and not exists (select r.* from registrations r where r.course = c.code and r.attendee = e.empno ) );
EMPNO ENAME INIT -------- -------- ----7499 ALLEN JAM SQL> Solution 9-4b. Using GROUP BY SQL> 2 3 4 5 6 7 8 9 10 11 12 e.empno, e.ename, e.init registrations r join courses c on (r.course = c.code) join employees e on (r.attendee = e.empno) where c.category = 'BLD' group by e.empno, e.ename, e.init having count(distinct r.course) = (select count(*) from courses where category = 'BLD'); select from
EMPNO ENAME INIT -------- -------- ----7499 ALLEN JAM SQL> This is not an easy exercise. Both of these solutions are correct. 5. Provide a list of all employees having the same monthly salary and commission as (at least) one employee of department 30. You are interested in only employees from other departments.
APPENDIX B
ANSWERS TO THE EXERCISES
Solution 9-5. SQL> 2 3 4 5 6 7 8 9 select e.ename , e.msal , e.comm from employees e where e.deptno <> 30 and ( e.msal,coalesce(e.comm,-1) ) in (select x.msal,coalesce(x.comm,-1) from employees x where x.deptno = 30 );
ENAME MSAL COMM -------- -------- -------SMITH 800 SQL> Note that this solution uses the COALESCE function, which you need to make comparisons with null values evaluate to true, in this case. The solution uses the value 1 based on the reasonable assumption that the commission column never contains negative values. However, if you check the definition of the EMPLOYEES table, you will see that there actually is no constraint to allow only nonnegative commission values. It looks like you found a possible data model enhancement here. Such a constraint would make your solution using the negative value in the COALESCE function correct under all circumstances. 6. Look again at Listings 9-4 and 9-5. Are they really logically equivalent Just for testing purposes, search on a nonexisting job and execute both queries again. Explain the results. Solution 9-6. SQL> select e.empno, e.ename, e.job, e.msal 2 from employees e 3 where e.msal > ALL (select b.msal 4 from employees b 5 where b.job = 'BARTENDER'); EMPNO -------7369 7499 7521 7566 7654 7698 7782 7788 7839 7844 7876 7900 ENAME -------SMITH ALLEN WARD JONES MARTIN BLAKE CLARK SCOTT KING TURNER ADAMS JONES JOB MSAL -------- -------TRAINER 800 SALESREP 1600 SALESREP 1250 MANAGER 2975 SALESREP 1250 MANAGER 2850 MANAGER 2450 TRAINER 3000 DIRECTOR 5000 SALESREP 1500 TRAINER 1100 ADMIN 800
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