List all the employees who are having reporting managers in dept 10 along with 10 hike in salary

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Very Important and basics SQL Queries for Testing: If you are looking for some solutions for the queries asked during the SQL round for developer and testing position, then you can check below a huge collection of SQL queries for the solution of the SQL DBMS queries related questions.

SQL Interview Questions

Display the details of all employees.
Ans: select * from emp;

Display the depart information from the department table.
Ans: select * from dept;

Display the name and job for all the employees.
Ans: select ename, job from emp;

Display the name and salary for all the employees.
Ans: select ename, sal from emp;

Display the employee no and the total salary for all the employees.
Ans: select empno, sal+comm as total from emp group by empno;

Display the employee name and annual salary for all employees.
Ans: select ename, sal * 12 as annual salary from emp;

Display the names of all the employees who are working in depart number 10.
Ans: select ename from emp where deptno=10;

Display the names of all the employees who are working as clerks and drawing a salary of more than 3000.
Ans: select ename from emp where job=’CLERKS ‘and sal>3000;

Display the employee number and name who are earning comm.
Ans: select empno, ename from emp where comm is not null;

Display the employee number and name who do not earn any comm.
Ans: select empno, ename from emp where comm is null;

Display the names of employees who are working as clerks, salesman, or analysts and drawing a salary of more than 3000.
Ans: select ename from emp where job=’CLERK’ OR JOB=’SALESMAN’ OR JOB=’ANALYST’ AND SAL>3000;

Display the names of the employees who are working in the company for the past 5 years.
Ans: select ename from emp where to_char (sysdate,’YYYY’) -to_char (hiredate,’YYYY’) >=5;

Display the list of employees who have joined the company before30-JUN-90 or after 31-DEC-90.
Ans: select ename from emp where hiredate < ’30- JUN-1990′ or hiredate >’31-DEC-90′;

Display current Date.
Ans: select sysdate from dual;

Display the list of all users in your database (use catalog table).
Ans: select username from all users;

Display the names of all tables from the current user.
Ans: select name from tab;

Display the name of the current user.
Ans: show user

SQL Query Interview Questions

Display the names of employees working in depart number 10 or 20 or 40 or employees working as CLERKS, SALESMAN, or ANALYST.
Ans: select ename from emp where deptno in (10, 20, 40) or job in (‘CLERKS’,’SALESMAN’,’ANALYST’);

Display the names of employees whose name starts with alphabet S.
Ans: select ename from emp where ename like ‘S%’;

Display the Employee names for employees whose name ends with Alphabet S.
Ans: select ename from emp where ename like’%S’;

Display the names of employees whose names have second alphabet A in their names.
Ans: Select ename from EMP where ename like ‘_A%’;

Select the names of the employee whose names are exactly five characters in length.
Ans: select ename from emp where length (ename) =5;

Display the names of the employee who are not working as MANAGERS.
Ans: select ename from emp where job not in (‘MANAGER’);

Display the names of the employee who are not working as SALESMAN OR CLERK OR ANALYST.
Ans: select ename from emp where job not in (‘SALESMAN’,’CLERK’,’ANALYST’);

Display all rows from the EMP table. The system should wait after every screen full of information.
Ans: Set pause on

Display the total number of employees working for the company.
Ans: Select count (*) from EMP;

Display the total salary begging paid to all employees.
Ans: select sum (sal) from emp;

Display the maximum salary from emp table.
Ans: select max (sal) from emp;

Display the minimum salary from emp table.
Ans: select min (sal) from emp;

Display the average salary from emp table.
Ans: select avg (sal) from emp;

Display the maximum salary being paid to the CLERK.
Ans: select max (sal) from emp where job=’CLERK’;

Display the maximum salary being paid to depart number 20.
Ans: select max (sal) from emp where deptno=20;

Display the minimum salary being paid to any SALESMAN.
Ans: select min (sal) from emp where job=’SALESMAN’;

Display the average salary drawn by MANAGERS.
Ans: select avg (sal) from emp where job=’MANAGER’;

Display the total salary drawn by ANALYST working in depart number40.
Ans: select sum (sal) from emp where job=’ANALYST’ and deptno=40;

Display the names of the employee in order of salary, i.e., the name of the employee earning the lowest salary should salary appear first.
Ans: select ename from emp order by sal;

Display the names of the employee in descending order of salary.
Ans: select ename from emp order by sal desc;

Display the names of the employee in order of employee name.
Ans: select ename from emp order by ename;

Display empno, ename, deptno, sal sort the output first base on the name and within name by deptno and within deptno by sal.
Ans: select empno, ename, deptno, sal from emp order by ename, deptno, sal;

Display the name of the employee along with their annual salary (sal*12).T the name of the employee earning the highest annual salary, should appear first.
Ans: select ename, sal*12 from emp order by sal desc;

Display name, salary, hra, pf, da, total salary for each employee. The output should be in the order of total salary; hra 15% of salary, da 10%of salary, pf 5%salary, and total salary will be (salary+hra+da-pf).
Ans: select ename, sal, sal/100*15 as hra, sal/100*5 as pf, sal/100*10 as da, sal+sal/100*15+sal/100*10-sal/100*5 as total from emp;

Display depart numbers and the total number of employees working in each department.
Ans: select deptno, count (deptno) from emp group by deptno;

Display the various jobs and the total number of employees within each job group.
Ans: select job, count (job) from emp group by job;

Display the depart numbers and total salary for each department.
Ans: select deptno, sum (sal) from emp group by deptno;

Display the depart numbers and max salary for each department.
Ans: select deptno, max (sal) from emp group by deptno;

Display the various jobs and total salary for each job.
Ans: select job, sum (sal) from emp group by job;

Display the various jobs and total salary for each job.
Ans: select job, min (sal) from EMP group by job;

Display the depart numbers with more than three employees in each dept.
Ans: select deptno, count (deptno) from emp group by deptno having count (*) >3;

Display the various jobs along with the total salary for each of the jobs where the total salary is greater than 40000.
Ans: select job, sum (sal) from emp group by job having sum (sal) >40000;

Display the various jobs along with the total number of employees in each job. The output should contain only those jobs with more than three employees.
Ans: select job, count (empno) from emp group by job having a count (job) >3

Display the name of the employee who earns the highest salary.
Ans: select ename from emp where sal= (select max (sal) from emp);

Display the employee number and name for an employee working as a clerk and earning the highest salary among clerks.
Ans: select empno, ename from emp where job=’CLERK’ and sal= (select max (sal) from emp where job=’CLERK’);

SQL Queries Interview Questions

Display the names of a salesman who earns the salary more than the highest salary of any clerk.
Ans: select ename, sal from emp where job=’SALESMAN’ and sal> (select max (sal) from emp where job=’CLERK’);

Display the names of clerks who earn a salary more than the lowest salary of any salesman.
Ans: select ename from emp where job=’CLERK’ and sal> (select min (sal) from emp where job=’SALESMAN’);

Display the names of employees who earn a salary more than that of Jones or that of salary greater than that of Scott.
Ans: select ename, sal from emp where sal> (select sal from emp where ename=’JONES’) and sal> (select sal from emp where ename=’SCOTT’);

Display the names of the employees who earn the highest salary in their respective departments.
Ans: select ename, sal, deptno from emp where sal in (select max (sal) from emp group by deptno);

Display the names of the employees who earn the highest salaries in their respective job groups.
Ans: select ename, sal, job from emp where sal in (select max (sal) from emp group by job)

Display the employee names who are working in the accounting department.
Ans: select ename from emp where deptno= (select deptno from dept where dname=’ACCOUNTING’)

Display the employee names who are working in Chicago.
Ans: select ename from emp where deptno= (select deptno from dept where LOC=’CHICAGO’)

Display the Job groups having a total salary greater than the maximum salary for managers.
Ans: SELECT JOB, SUM (SAL) FROM EMP GROUP BY JOB HAVING SUM (SAL) > (SELECT MAX (SAL) FROM EMP WHERE JOB=’MANAGER’);

Display the names of employees from department number 10 with a salary greater than that of any employee working in other departments.
Ans: select ename from emp where deptno=10 and sal>any (select sal from emp where deptno not in 10).

Display the names of the employees from department number 10 with a salary greater than that of all employees working in other departments.
Ans: select ename from emp where deptno=10 and sal>all (select sal from emp where deptno not in 10).

Display the names of the employees in Uppercase.
Ans: select upper (ename) from emp;

Display the names of the employees in Lowercase.
Ans: select lower (ename) from emp;

Display the names of the employees in Proper case.
Ans: select initcap (ename) from emp;

Display the length of your name using an appropriate function.
Ans: select length (‘name’) from dual;

Display the length of all the employee names.
Ans: select length (ename) from emp;

Select the name of the employee to concatenate with the employee number.
Ans: select ename||empno from emp;

User appropriate function and extract 3characters starting from 2characters from the following string ‘Oracle.’ I.e., the output should be ‘ac.’
Ans: select substr (‘oracle’, 3, 2) from dual

find the First occurrence of the character ‘a’ from the following string, i.e., ‘Computer Maintenance Corporation.’
Ans: SELECT INSTR (‘Computer Maintenance Corporation’,’ a’, 1) FROM DUAL

Replace every occurrence of alphabet A with B in the string Allens (use translate function)
Ans: select translate (‘Allens’,’A’,’B’) from dual

Display the information from the emp table. Where the job manager is found, it should be displayed as boos (Use replace function).
Ans: select replace (JOB,’MANAGER’,’BOSS’) FROM EMP;

Display empno, ename, deptno from emp table. Instead of a display, department numbers display the related department name (Use decode function).
Ans: select empno, ename, decode (deptno, 10,’ACCOUNTING’, 20,’RESEARCH’, 30,’SALES’, 40,’OPRATIONS’) from emp;

Display your age in days.
Ans: select to date (sysdate) -to date (’10-Sep-77′) from dual

Display your age in months.
Ans: select months between (sysdate, ’10-Sep-77′) from dual

Display the current date as 15th August Friday Nineteen Ninety Seven.
Ans: select to char (sysdate,’ddth Month day year’) from dual

Scott has joined the company on Wednesday, 13th August, nineteen ninety.
Ans: select ENAME||’ HAS JOINED THE COMPANY ON’||to_char (HIREDATE,’dayddth Month year’) from EMP;

Find the date for the nearest Saturday after the current date.
Ans: SELECT NEXT_DAY (SYSDATE, ‘SATURDAY’) FROM DUAL;

Display the current time.
Ans: select to_char (sysdate,’hh:MM:ss’) from dual.

Display the date three months before the current date.
Ans: select add months (sysdate, 3) from dual;

Display the common jobs from department number 10 and 20.
Ans: select job from emp where deptno=10 and job in (select job from emp where deptno=20);

Display the jobs found in department 10 and 20 Eliminate duplicate jobs.
Ans: select distinct (job) from emp where deptno=10or deptno=20orselect distinct (job) from emp where deptno in (10, 20);

Display the jobs which are unique to department 10.
Ans: select distinct (job) from emp where deptno=10

Display the details of those who do not have any person working under them.
Ans: select e.ename from emp, emp e whereemp.mgr=e.empno group by e.ename having count (*) =1;

Display the details of those employees who are in the sales department and grade is 3.
Ans: select * from emp where deptno= (select deptno from dept where dname=’SALES’) and sal between (select low sal from salgrade where grade=3) and (select hisal from salgrade where grade=3);

Display those who are not managers and who are managers, anyone. i) Display the managers’ names.

Ans: select distinct (m.ename) from emp e, emp mwhere m.empno=e.mgr;

ii) Display the who are not managers.

Ans: select ename from emp where ename not in (select distinct (m.ename) from emp e, emp m where m.empno=e.mgr);

Display those employees whose name contains not less than 4characters.
Ans: select ename from emp where length (ename) >4;

Display that department whose name starts with “S” while the location name ends with “K”.
Ans: select dname from dept where dname like ‘S%’and loc like ‘%K’;

Display those employees whose manager’s name is JONES.
Ans: select p.ename from emp e, emp p where e.empno=p.mgr and e.ename=’JONES’;

Display those employees whose salary is more than 3000 after giving20% increment.
Ans: select ename, sal from emp where (sal+sal*.2) >3000;

Display all employees while their dept names;
Ans: select ename, dname from emp, dept where emp.deptno=dept.deptno

Display ename who are working in the sales dept.
Ans: select ename from emp where deptno= (select deptno from dept where dname=’SALES’);

Display employee name, deptname, salary, and comm. For those Sal in between2000 to 5000 while the location in Chicago.
Ans: select empno, ename, deptno from emp where deptno= (select deptno from dept where loc=’CHICAGO’) and sal between 2000 and 5000;

Display those employees whose salary greater than his manager’s salary.
Ans: select * from emp e where sal> (select sal from emp where empno=e.mgr);

Display those employees who are working in the same Dept where his manager is working.
Ans: select * from emp e where deptno = (select deptno from emp where empno=e.mgr);

Display those employees who are not working under any manger.
Ans: select * from emp where mgr is null or empno=mgr;

Display grade and employee name for the dept no 10 or 30, but the grade is not 4, while joined the company before 31-Dec-82.
Ans: select empno, ename, sal, deptno, hiredate, grade from emp e, salgrade where e.sal>=s.losal and e.sal<=s.hisal and deptno in (10, 30) andgrade<>4 and hiredate<’01-Dec-1981′;

Update the salary of each employee by 10% increments that are not eligible for commission.
Ans: update emp set sal=sal+ (sal*10/100) where comm is null;

delete those employees who joined the company before 31-Dec-82 while there dept location is ‘NEW YORK’ or ‘CHICAGO.’
Ans: delete from emp where hiredate<’31-Dec-1982′ and deptno in (select deptno from dept where loc in (‘NEW YORK’,’CHICAGO’));

Display employee name, job, deptname, location for all who are working as managers.
Ans: select ename, job, dname, loc from emp e, dept d where e.deptno=d.deptno and empno in (select mgr from emp);

Display those employees whose manager names are Jones, and also display their manager name.
Ans: select e.empno, e.ename, m.ename MANAGER from emp e, emp where e.mgr=m.empno and m.ename=’JONES’;

Display name and salary of ford if his Sal is equal to high Sal of his grade.
Ans: select ename, sal from emp e where ename=’FORD’ and sal=(select hisal from salgrade where grade=(select grade from salgrade where e.sal>=losal and e.sal<=hisal));

Display employee name, his job, his dept name, his manager name, his grade and make out of an under department wise.break on deptno;
Ans: select d.deptno, e.ename, e.job, d.dname, m.ename, s.grade from emp e, emp m, Dept d, salgrade s where e.deptno=d.deptno and e.salbetween s.losal and s.hisal and e.mgr=m.empno order by e.deptno;

List out all the employee’s name, job, and salary grade and department name for everyone in the company except ‘CLERK’. Sort on salary displays the highest salary.
Ans: select empno, ename, sal, dname, grade from emp e, dept d, salgrade where e.deptno=d.deptno and e.sal betweens.losal and s.hisal ande.job<>’CLERK’ order by sal;

Display employee name, his job, and his manager. Display also employees’ who are without the manager.
Ans: select e.ename, e.job, m.ename Manager from emp e,emp m where.mgr=m.empno union select ename, job,’ no manager’ from emp where mgr is null;

SQL Interview Questions And Answers

Find out the top 5 earners of the company.
Ans: select * from emp e where 5> (select count (*) from emp wheresal>e.sal) order by sal desc;

Display the name of those employees who are getting the highest salary.
Ans: select empno, ename, sal from emp where sal=(select max(sal) from emp);

Display those employees whose salary is equal to an average of maximum and minimum.
Ans: select * from emp where sal= (select (max (sal) +min (sal))/2 fromemp);

Display count of employees in each department where count greater than 3.
Ans: select deptno, count (*) from emp group by deptno having count (*)>3;

Display dname where at least 3 are working and display the only dname.
Ans: select dname from dept where deptno in (select deptno from emp group by deptno having count (*)>3);

Display the name of those managers name whose salary is more than the average salary of a company.
Ans: select ename, sal from emp where empno in (select mgr from emp) and sal > (select avg (sal) from emp);

Display those manager’s names whose salary is more than the average salary of his employees.
Ans: select ename, sal from emp e where empno in (select mgr from emp) and e.sal> (select avg (sal) from emp where mgr=e.empno);

Display employee name, Sal, commission, and net pay for those employees whose net pay are greater than or equal to any other employee salary of the company?
Ans: select ename, sal, comm, sal+nvl (comm, 0) netPay from emp wheresal+nvl (comm., 0)>=any (select sal from emp);

Display those employees whose salary is less than his manager but more than the salary of any other managers.
Ans: select * from emp e where sal<(select sal from emp where empno =e.mgr) and sal>any(select sal from emp where empno!=e.mgr);

Display all employee’s names with total Sal of a company with each employee name.
Ans: Select ename, (select sum (sal) from emp) from emp;

Find out the last 5(least) earner of the company?
Ans: select * from emp e where 5> (select count (*) from emp wheresal<e.sal) order by sal;

Find out the number of employees whose salary is greater than there manager salary?
Ans: select count (*) from emp e where sal> (select sal from emp whereempno=e.mgr);

Display that manager who is not working under the president, but they are working under any other manager?
Ans: select * from emp e where mgr in (select empno from emp whereename<> ‘KING’);

Delete those departments where is no employee working?
Ans: delete from dept d where 0= (select count (*) from emp wheredeptno=d.deptno);

Delete those records from the EMP table whose deptno not available in the dept table?
Ans: delete from emp where deptno not in (select deptno from dept);

Display those earners whose salary is out of the grade available in the Sal grade table?
Ans: select * from emp where sal< (select min (losal) from salgrade) orsal> (select max (hisal) from salgrade);

Display employee name, Sal, commission. and whose net pay is greater than any other in the company?
Ans: select ename, sal, comm from emp where sal+sal*15/100-sal*5/100+sal*10/100 = (select max (sal+sal*15/100-sal*5/100+sal*10/100) from emp);

Display name of those employees who are going to retire 31-Dec-99. If the maximum job is period is 18 years?
Ans: select * from emp where (to_date (’31-dec-1999′)-hiredate)/365>18;

Display those employees whose salary is ODD value?
Ans: select * from emp where mod (sal, 2) =1;

Display those employees whose salary contains at least 4 digits?
Ans: select * from emp where length (sal)>=4;

Display those employees who joined the company in the month of DEC?
Ans: select * from emp where upper (to_char (hiredate,’mon’)) =’DEC’;

Display those employees whose name contains “A”?
Ans: select * from emp where instr (ename,’A’, 1, 1)>0;

Display those employees whose deptno is available in salary?
Ans: select * from emp where instr (sal, deptno, 1, 1)>0;

Display those employees whose first 2 characters from hire date-last 2characters of salary?
Ans: select substr(hiredate,0,2)||substr(sal,length(sal)-1,2) from emp; select concat( substr(hiredate,0,2), substr(sal,length(sal)-1,2) ) fromemp;

Display those employees whose 10% of salary is equal to the year of joining?
Ans: select * from emp where to_char (hiredate,yy’) =sal*10/100;

Display those employees who are working in sales or research?
Ans: select * from emp where deptno in (select deptno from dept where name in (‘SALES’,’RESEARCH’));

Display the grade of Jones?
Ans: select grade from salgrade where losal<= (select (sal) from emp whereename=’JONES’) and hisal>= (select (sal) from emp whereename=’JONES’);

Display those employees who joined the company before the 15Th of the month?
Ans: select empno, ename from emp where hiredate< (to_date (’15-‘||to_char(hiredate,’mon’)||’-‘||to_char(hiredate,’yyyy’)));

Delete those records where no of employee in a particular department is lessthan 3?
Ans: delete from emp where deptno in (select deptno from emp group by deptno having count (*)>3);

Delete those employees who joined the company 21 years back from today?
Ans: select * from emp where round ((sysdate-hiredate)/365)>21; Or

Ans: select * from emp where (to_char (sysdate, ‘yyyy’)-to_char (hiredate,’yyyy’))>21;

Display the department name the no of characters of which is equal to no of employees in any other department?
Ans: select dname from dept where length (dname) in (select count (*) from emp group by deptno);

Display those employees who are working as a manager?
Ans: select * from emp where empno in (select mgr from emp);

Count the no of employees who are working as a manager (use set operation)?
Ans: select count (*) from emp where empno in (select mgr from emp);

Display the name of then dept those employees who joined the company on the same date?
Ans: select empno, ename, hiredate, deptno from emp e where hiredate in (select hiredate from emp where empno<>e.empno);

Display the manager who is having a maximum number of employees working under him?
Ans: select mgr from emp group by mgr having count (*) =(select max(count(mgr)) from emp group by mgr);

List out employee’s name, and salary increased by 15% and expressed as the whole number of dollars?
Ans: select empno, ename, lpad (concat (‘$’, round (sal*115/100)), 7) salary from emp;

Produce the output of the EMP table “EMPLOYEE_AND_JOB” forename and job?
Ans: select * from EMPLOYEE_AND_JOB;

List all employees with hire date in the format ‘June 4, 1988’?
Ans: select to_char (hiredate,’ month dd yyyy’) from emp;

Print a list of employees displaying ‘Less Salary’ if less than 1500 if exactly1500 display as ‘Exact Salary’ and if greater than 1500 display ‘More Salary’?
Ans: select empno, ename,’Less Salary ‘||sal from emp where sal<1500union select empno, ename,’More Salary ‘||sal from emp where sal>1500union select empno, ename,’Exact Salary ‘||sal from emp where sal=1500

Write a query to calculate the length of the employee who has been with the company?
Ans: select round (sysdate-hiredate) from emp;

Display those managers who are getting less than their employees Sal.
Ans: select empno from emp e where sal<any (select sal from emp wheremgr=e.empno);

Print the details of all the employees who are subordinate to Blake.
Ans: select * from emp where mgr= (select empno from emp whereename=’BLAKE’);

Display those who work as a manager using a co-related subquery.
Ans: select * from emp where empno in (select mgr from emp);

Display those employees whose manager’s name is Jones and also with his manager’s name.
Ans: select * from emp where mgr= (select empno from emp whereename=’JONES’) union select * from emp where empno= (select mgrfrom emp where ename=’JONES’);

Use the variable in a statement which finds all employees who can earn 30,000a year or more.
Ans: select * from emp where &emp_ann_sal>30000;

Find out how many managers are there without listing them.
Ans: select count (*) from EMP where empno in (select mgr from EMP);

Find out the avg sal and avg total remuneration for each job type remember salesmen earn the commission.
Ans: select job, avg (sal+nvl (comm, 0)), sum (sal+nvl (comm, 0)) from empgroup by job;

Check whether all employees number are indeed unique.
Ans: select count(empno),count(distinct(empno)) from emp havingcount(empno)=(count(distinct(empno)));

List out the lowest paid employees working for each manager, exclude anygroups where min sal is less than 1000 sort the output by sal.
Ans: select e.ename, e.mgr, e.sal from emp e where sal in (select min (sal) from emp where mgr=e.mgr) and e.sal>1000 order by sal;

list ename, job, annual salary, deptno, dname, and grade who earn 30000 per year and who are not clerks.
Ans: select e.ename, e.job, (e.sal+nvl (e.comm,0))*12, e.deptno, d.dname,s.grade from emp e, salgrade s , dept d where e.sal between s.losaland s.hisal and e.deptno=d.deptno and (e.sal+nvl(comm,0))*12>30000 and e.job <> ‘CLERK’;

find out the all employees who joined the company before their manager.
Ans: select * from emp e where hiredate< (select hiredate from emp whereempno=e.mgr);

list out all employees by name and number along with their manager’s name and number also display ‘No Manager’ who has no manager.
Ans: select e.empno, e.ename, m.empno Manager, m.ename ManagerNamefrom emp e, emp m where e.mgr=m.emp union select empno, ename, mgr,’No Manager’ from emp where mgr is null;

find out the employees who earned the highest Sal in each job typed sort in descending Sal order.
Ans: select * from emp e where sal = (select max (sal) from emp where job=e.job);

find out the employees who earned the min Sal for their job in ascending order.
Ans: select * from emp e where sal = (select min (sal) from emp where job=e.job) order by sal;

find out the most recently hired employees in each dept order by hire date.
Ans: select * from emp order by deptno, hiredate desc;

display ename, sal, and deptno for each employee who earns a Sal greater than the avg of their department order by deptno.
Ans: select ename, sal, deptno from emp e where sal> (select avg (sal) from emp where deptno=e.deptno) order by deptno;

SQL Interview Questions For Experienced

display the department where there are no employees.
Ans: select deptno, dname from dept where deptno not in (select distinct(deptno) from emp);

Display the dept no with the highest annual remuneration bill as compensation.
Ans: select deptno, sum (sal) from emp group by deptno having sum (sal) =(select max(sum(sal)) from emp group by deptno);

In which year did most people join the company. Display the year and number of employees.
Ans: select count (*), to_char (hiredate,’yyyy’) from emp group byto_char (hiredate,’yyyy’);

display avg sal figure for the dept.
Ans: select deptno, avg (sal) from emp group by deptno;

Write a query of display against the row of the most recently hired employee. Display ename hire date and column max date showing.
Ans: select empno, hiredate from emp where hiredate= (select max (hiredate) from emp);

display employees who can earn more than the lowest Sal in dept no 30.
Ans: select * from emp where sal> (select min (sal) from emp wheredeptno=30);

find employees who can earn more than every employee in dept no 30.
Ans: select * from emp where sal> (select max (sal) from emp wheredeptno=30); select * from emp where sal>all (select sal from emp wheredeptno=30);
select dept name dept no and the sum of Sal break on deptno on dname.
Ans: select e.deptno, d.dname, sal from emp e, dept d where e.deptno=d.deptno order by e.deptno;

find all dept’s which have more than 3 employees.
Ans: select deptno from emp group by deptno having count (*)>3;

Display half of the enames in the upper case and the remaining lower case.
Ans: select concat ( upper ( substr ( ename, 0 , length (ename)/ 2) ),lower (substr (ename, length(ename) / 2+1, length(ename) )) ) from emp;

Select ename if ename exists more than once.
Ans: select distinct (ename) from emp e where ename in (select ename from emp where e.empno<>empno);

display all enames in reverse order.
Ans: select ename from emp order by ename desc;

Display that employee whose joining of month and grade is equal.
Ans: select empno, ename from emp e, salgrade s where e.sal betweens.losal and s.hisal and to_char (hiredate,’ mm’) =grade;

Display that employee whose joining date is available in dept no.
Ans: select * from emp where to_char (hiredate,’dd’) =deptno;

Display those employee’s names as follows A ALLEN, B BLAKE.
Ans: select substr (ename, 1, 1) ||’ ‘||ename from emp;

List out the employees ename, sal, PF from emp.
Ans: select ename, sal, sal*15/100 PF from emp;

Create table emp with only one column empno.
Ans: Create table emp (empno number (5));

Add this column to emp table ename Varchar (20).
Ans: alter table emp add ename varchar2 (20) not null;

OOPS! I forgot to give the primary key constraint. Add it now.
Ans: alter table emp add constraint emp_empno primary key (empno);

now increase the length of the ename column to 30 characters.
Ans: alter table emp modify ename varchar2 (30);

Add salary column to emp table.
Ans: alter table emp add sal number (7, 2);

I want to give a validation saying that sal cannot be greater 10,000(note provide a name for this column).
Ans: alter table emp add constraint emp_sal_check check (sal<10000);

For the time being, I have decided that I will not impose this validation. My boss has agreed to pay more than 10,000.
Ans: Alter table emp disable constraint emp_sal_check;

My boss has changed his mind. Now he doesn’t want to pay more than10, 000. So revoke that salary constraint
Ans: Alter table emp enable constraint emp_sal_check;

Add a column called mgr to your emp table.
Ans: Alter table emp add mgr number (5);

Oh! This column should be related to the empno. Give a command to add this constraint.
Ans: Alter table emp add constraint emp_mgr foreign key (empno);

Add dept no column to your emp table.
Ans: Alter table emp add deptno number (3);

This dept no column should be related to a deptno column of the dept table.
Ans: Alter table emp1 add constraint emp1_deptno foreign key (deptno) references dept (deptno);

Create a table called newemp. Using a single command creates this table as well as to get data into this table (use create table as).
Ans: create table newemp as select *from emp;

Create a table called newemp. This table should contain only empno, ename, dname.
Ans: create table newemp as select empno, ename, dname from emp e, dept d where e.deptno=d.deptno;

Delete the rows of employees who are working in the company for more than 2years.
Ans: Delete from emp where floor (sysdate-hiredate)>2*365;

Provide a commission to employees who are not earning any commission.
Ans: update emp set comm=300 where comm is null;

If any employee has commissioned his commission should be incremented by10% of his salary.
Ans: update emp set comm=comm*10/100 where comm is not null;

Display employee name and department name for each employee.
Ans: select ename, dname from emp e, dept d where e.deptno=d.deptno;

Display employee number, name, and location of the department in which he is working.
Ans: select empno, ename, loc from emp e, dept d where e.deptno=d.deptno;

Display ename, dname even if there no employees working in a particular department (use outer join).
Ans: select ename, dname from emp e, dept d where e.deptno (+) =d.deptno;

Display employee name and his manager’s name.
Ans: select e.ename, m.ename from emp e, emp m where e.mgr=m.empno;

Display the department name along with the total salary in each department.
Ans: select deptno, sum (sal) from emp group by deptno;

Display the department name and the total number of employees in each department.
Ans: select deptno, count (*) from emp group by deptno;

Display the current date and time.
Ans: select to_char (sysdate,’month mon dd yy yyyy hh: mm: ss’) from dual;

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