Ashutosh Srivastava

• Sep 21st, 2007

 

select * from emp where sal in(
select sal from emp where rowid not in(
select max(rowid) from emp group by sal))

vanigreat

 Profile Answers by vanigreat 

• Jan 8th, 2008

 

select a.empno,a.ename,a.sal from scott.emp a where a.sal in

(select sal from scott.emp group by sal having count(*)>1)

Shilpa.Paschapur

 Profile Answers by Shilpa.Paschapur 

• Jan 22nd, 2008

 

 hi..
select name,salary from emp e1,emp e2 where e1.esalary=e2.esalary

patilpravin_1981

 Profile Answers by patilpravin_1981 

• Jan 29th, 2008

 

select e1.ename,e2.salary from emp e1.emp e2 where
e1.emp_id=e2.emp_id

sureshkumar.mtech

 Profile Answers by sureshkumar.mtech Questions by sureshkumar.mtech

• Oct 15th, 2008

 

SQL> select distinct e.name,e.salary from employe e, employe a
  2  where e.salary = a.salary and e.name != a.name;

NAME           SALARY
---------- ----------
MANI            10000
SELVAM          10000
SURAJ           10000
KAMAL           20000
RAMESH          20000
SARA            20000

6 rows selected.

kethinenisarath

 Profile Answers by kethinenisarath 

• Mar 8th, 2011

 

This query display the same salary.
 
select  *  from  emp
where sal in (select sal from emp
group by sal
having count(1)>1)



Plz let me know in case of any problem

Thanks
Sarath

Prabakaran

• Jul 27th, 2011

 

select * from emp where eid in (select eid from emp groupby emp.sal having count(*)>1)

vishnu

• Aug 7th, 2011

 

select e.ename,e.salary from emp e self join emp s on e.salary =s.salary

jo

• Aug 9th, 2011

 

select ename,sal from emp where sal in(select sal from emp group by sal having count(sal)>=2)

fire_starter

 Profile Answers by fire_starter 

• Oct 9th, 2011

 

Hello,

try this:

select emp_id, name, salary, rank() over ( order by salary ) from employee;

leelakrishna302

 Profile Answers by leelakrishna302 

• Jun 15th, 2012

 

select distinct D. Ename,D.SAL from EMP E,EMP D WHERE E.SAL = D.SAL AND E.ENAME <> D.ENAME;

honeyanubha

 Profile Answers by honeyanubha 

• Jun 22nd, 2012

 

I think the easiest way is to SELECT A.* FROM EMPLOYEE A, EMPLOYEE B WHERE A.SAL = B.SAl AND A.ROWID <> B.ROWID

Code
SELECT A.* FROM EMPLOYEE A, EMPLOYEE B WHERE A.SAL = B.SAl AND A.ROWID <> B.ROWID

vikash kumar singh

• Aug 4th, 2012

 

Code
SELECT eMployee_Id,First_Name,Last_Name,Salary FROM(SELECT Employee_Id,First_Name,Last_Name,Salary,Count(*) Over (Partition BY Salary ORDER BY Salary) Cnt 


FROM Employees) WHERE Cnt>=2;

manjari_ghosh12

 Profile Answers by manjari_ghosh12 

• Sep 1st, 2012

 

Code
SELECT e1.ename,e1.sal,count(*) person_having_same_sal FROM emp e1,emp e2 WHERE e1.sal=e2.sal GROUP BY e1.sal,e1.ename HAVING count(*)>1;

OR

Code
SELECT * FROM emp WHERE sal IN (SELECT sal FROM emp GROUP BY sal HAVING count(*)>1);

OR

Code
SELECT DISTINCT e1.ename,e1.sal FROM emp e1,emp e2 WHERE e1.sal=e2.sal AND e1.ename!=e2.ename;

Revel

• Aug 24th, 2016

 

Why we used count(1)>1 here? Can you explain please?

amaresh

• Mar 16th, 2017

 

Assuming your question is why the guy has used count(*)>1, the reason is: First we need to group by all the values, when you group by the result set is all the values which are having same values w.r.to that coulmn will be under single group. Suppose two employees having same salary as 5000, then under 5000 group we have two employees and in order to find if the employees are more than one or not we have to use count(*)>1 ...simple ...enjoy

Pinky

• May 31st, 2017

 

SELECT * FROM EMPLOYEE E1, EMPLOYEE E2 WHERE E.SAL = E2.SAL AND E1.EMP_ID <> E2.EMP_ID

Ranaj Kumar Parida

• Jul 31st, 2017

 

SELECT ENAME,JOB SAL,DEPTNO FROM EMP
WHERE SAL IN
(
SELECT SAL FROM EMP
GROUP BY SAL
HAVING COUNT(*)>1
);

amit

• Oct 20th, 2017

 

SELECT SAL,listagg(ename,,) within group(order by ename) same_sal from t1
group by sal
having count(ename)>1;

Asish

• Jan 7th, 2018

 

Select e.* From employee e, employee f
Where e.employee_id != f.employee_id
And e.sal = f.sal

RANAJ KUMAR PARIDA

• Feb 27th, 2018

 

SELECT E1.* FROM EMP E1
WHERE SAL
IN
(
SELECT SAL FROM EMP
GROUP BY SAL
HAVING COUNT(*) > 1
);

Ekansh

• Jul 12th, 2022

 

just add the ; in the end of quest :)

Vamsikrishna Sudanagunta

• Nov 16th, 2022

 

Print first names and the last names of the artists who had zero sales or no sales?