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Answers for SQL Functions



1. SQL> SELECT empno, ename FROM emp WHERE Length(ename) = 4;

2. SQL> SELECT empno, ename, job FROM emp where Length(job)=7;

3. SQL> SELECT Length('qspiders') - Length(replace('qspiders','s','')) FROM dual;

4. SQL>  SELECT empno, ename, job FROM emp WHERE Instr(job,'MAN') >0;

5. SQL> SELECT empno, ename, job FROM emp WHERE Instr(job, 'MAN') =1;

6. SQL> SELECT empno, ename, job FROM emp WHERE (Length(ename) - Length(Replace(ename, 'L',''))) = 1;

7. SQL> SELECT * FROM dept WHERE Instr(dname,'O') > 0;

8. SQL> SELECT Concat(ename,' working as a ') || Concat(job, ' earns ') || Concat(sal, '  in ') || Conc
at('dept ',deptno) AS text from emp;

OR

SQL> SELECT Concat(Concat(Concat(Concat(Concat(Concat(Concat(ename,' working as a '), job),' earns '), sal),'  in '),'dept '), deptno) AS text FROM emp;

9. SQL> SELECT empno, ename, sal FROM emp WHERE mod(sal,2) > 0;


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Comments

Rakesh said…
sir can u pls add all the answers for functions (only 9 ans are there)
and also answers for qspider questions,,,,,
November 4, 2011 at 3:32 PM
Shailaja Yadav said…
Sir, can you plz add all the answers,there are only 9 answers...
November 24, 2011 at 1:05 PM
R.N SOUMYA RANJAN SAMAL said…
I AM NOT GETTING THE EXACT O/P FROM THE ANSWER NO-4.ACTUALLY I AM GETTING ALL THE MANAGER AND SALESMAN ON MY SCREEN.SO KINDLY IF YOU MODIFY I'LL BE THANKFUL.
August 3, 2012 at 4:55 PM
Rekhamopuru said…
select * from employees where instr(job_id,'MAN')=(length(job_id)-length('MAN')+1);
i hope this will work for the 4th question
August 16, 2012 at 7:14 AM
Rekhamopuru said…
i hope this will give better output for 4th question
August 16, 2012 at 7:15 AM
Soumya said…
Dear rekhamopuru
for the 4th question how could you find the job_id from emp table
there is no job_id in EMP table.hope for the next answer
August 16, 2012 at 10:06 PM
Rekhamopuru said…
Soumya...in my table i named column as job_id..u can give u r own name like job..it dont nt matter ....
u try job insted of job_id...
August 16, 2012 at 10:10 PM
jeremia said…
sir,i think your answer to Q-4 is wrong because it will display all the employees whose job are having 'MAN' as the last three character or as the first three character or any other position in the string...so sir, the answer should be as follow:-
select * from empwhere substr(job,-3,3)='MAN';
August 30, 2012 at 7:48 PM
jeremia said…
answer to Q-10
select count(empno) from emp where comm is null;
August 30, 2012 at 10:23 PM
jeremia said…
answer to Q-11
select sum(sal),sum(comm) from emp where deptno=30;
August 30, 2012 at 10:27 PM
jeremia said…
answer to Q-12
select deptno,count(empno) no_of_clerks from emp where job='CLERK' and deptno in(20,30) group by deptno;
August 30, 2012 at 10:34 PM
jeremia said…
answer to Q-13
select deptno,sum(sal) total_salary from emp group by deptno;
August 30, 2012 at 10:38 PM
jeremia said…
answer to Q-14
select deptno,sum(sal) total_salary from emp
group by deptno
having sum(sal)>3000;
August 30, 2012 at 10:40 PM
jeremia said…
answer to Q-15
select job,sum(sal) total_sal from emp where deptno not in 30 group by job having sum(sal)>5000;
August 30, 2012 at 11:18 PM
jeremia said…
answer to Q-16
select job,max(sal) from emp where deptno in(10,20) and job in('MANAGER','CLERK','SALESMAN') group by job order by max(sal) desc;
August 30, 2012 at 11:42 PM
jeremia said…
answer to Q-17
select job,sum(sal) total_salary,count(empno) no_of_employees from emp group by job having sum(sal)>5000;
August 31, 2012 at 12:28 AM
jeremia said…
correction in answer of Q-17.sum(sal)>9000;replace 5000 with 9000.
August 31, 2012 at 12:30 AM
jeremia said…
answer to Q-18
select deptno,count(empno) no_of_employees from emp group by deptno having count(empno)>=4;
August 31, 2012 at 12:38 AM
jeremia said…
answer to Q-19
select distinct deptno from emp where deptno in (select deptno from emp group by deptno having count(distinct job)=1) and job='SALESMAN';
August 31, 2012 at 3:07 AM
jeremia said…
answer to Q-20
select length('HELLLLL')-length(replace('HELLLLL','L')) no_of_L from dual;
August 31, 2012 at 3:17 AM
jeremia said…
answer to Q-21
select * from emp where instr(job,'MAN',1)>0;
August 31, 2012 at 4:38 PM
jeremia said…
answer to Q-22
select * from emp where substr(job,1,3)='MAN';
August 31, 2012 at 4:40 PM
jeremia said…
answer to Q-23
select * from emp where substr(job,-3,3)='MAN';
August 31, 2012 at 4:42 PM
jeremia said…
answer to Q-25
select ename,lower(substr(ename,1,3))||upper(substr(ename,4)) from emp;
August 31, 2012 at 4:48 PM
jeremia said…
answer to Q-26
select ename || 'is a' || job || 'and gets salary' || sal from emp;

(or)
you can also use the word "concat"
August 31, 2012 at 4:51 PM
MAYANK said…
SIR IF WE ARE SPECIFYING INSTR(JOB,'MAN',1)>0; THEN WE ARE GETTING THE RECORDS HAVING CHARACTERS MAN EITHER IN STARTING OR END POSITION OF THE JOB BUT ACCORDING TO THE QUESTION IT SHOULD DISPLAY THE EMPLOYEES HAVING JOB WITH THE STARTING STRING 'MAN' TO GET THE RECORDS ACCORDINGLY WE NEED TO WRITE INSTR(JOB,'MAN',1)=1 WHICH LEADS TO THE EXACT SOLUTION OF THE GIVEN QUESTION...
THANKS
September 24, 2012 at 9:30 PM
Basavanagowda K G said…
This comment has been removed by the author.
September 8, 2014 at 8:35 PM
Basavanagowda K G said…
hi sir for Question NO. 4
SELECT * FROM EMP
WHERE SUBSTR(JOB,1,3)='MAN';
--will select all the employees who are mangers
SELECT * FROM EMP
WHERE SUBSTR(JOB,1)='MAN';
--says no rows selected
why is it so
September 8, 2014 at 8:38 PM
Ravichandra Maddineni said…
sir please provide all answers
October 22, 2014 at 12:37 PM
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