SQL Interview Questions part 6
You can see part 5 here.
76) Display the following output for each row from emp table.
Scott has joined the company on Wednesday 13th August nineteen ninety.
[hideshow type=”sql76″]SQL>select ENAME||’ HAS JOINED THE COMPANY ON ‘||to_char(HIREDATE,’day ddth Month year’) from EMP;[/hideshow]
77) Find the date for nearest saturday after current date.
[hideshow type=”sql77″]SQL>SELECT NEXT_DAY(SYSDATE,’SATURDAY’)FROM DUAL;[/hideshow]
78) Display current time.
[hideshow type=”sql78″]SQL>select to_char(sysdate,’hh:MM:ss’) from dual;[/hideshow]
79) Display the date three months Before the current date.
[hideshow type=”sql79″]SQL>select add_months(sysdate,3) from dual;[/hideshow]
80) Display the common jobs from department number 10 and 20.
[hideshow type=”sql80″]SQL>select job from emp where deptno=10 and job in(select job from emp where deptno=20);[/hideshow]
81) Display the jobs found in department 10 and 20 Eliminate duplicate jobs.
[hideshow type=”sql81″]
SQL>select distinct(job) from emp where deptno=10 or deptno=20
(or)
SQL>select distinct(job) from emp where deptno in(10,20);
[/hideshow]
82) Display the jobs which are unique to department 10.
[hideshow type=”sql82″]SQL>select distinct(job) from emp where deptno=10;[/hideshow]
83) Display the details of those who do not have any person working under them.
[hideshow type=”sql83″]SQL>select e.ename from emp,emp e where emp.mgr=e.empno group by e.ename having count(*)=1;[/hideshow]
84) Display the details of those employees who are in sales department and grade is 3.
[hideshow type=”sql84″]SQL>select * from emp where deptno=(select deptno from dept where dname=’SALES’)and sal between(select losal from salgrade where grade=3)and (select hisal from salgrade where grade=3);[/hideshow]
85) Display those who are not managers and who are managers any one.
i)display the managers names
[hideshow type=”sql851″]SQL>select distinct(m.ename) from emp e,emp m where m.empno=e.mgr;[/hideshow]
ii)display the who are not managers
[hideshow type=”sql852″]SQL>select ename from emp where ename not in(select distinct(m.ename) from emp e,emp m where m.empno=e.mgr);[/hideshow]
86) Display those employee whose name contains not less than 4 characters.
[hideshow type=”sql86″]SQL>select ename from emp where length(ename)>4;[/hideshow]
87) Display those department whose name start with “S” while the location name ends with “K”.
[hideshow type=”sql87″]SQL>select dname from dept where dname like ‘S%’ and loc like ‘%K’;[/hideshow]
88) Display those employees whose manager name is JONES.
[hideshow type=”sql88″]SQL>select p.ename from emp e,emp p where e.empno=p.mgr and e.ename=’JONES’;[/hideshow]
89) Display those employees whose salary is more than 3000 after giving 20% increment.
[hideshow type=”sql89″]SQL>select ename,sal from emp where (sal+sal*.2)>3000;[/hideshow]
90) Display all employees while their dept names.
[hideshow type=”sql90″]SQL>select ename,dname from emp,dept where emp.deptno=dept.deptno;[/hideshow]
