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You have one employee table having 3 entries emp_id, Mgr_id
and salary. Mgr_id is also referencing emp_id. so now find
out the emp_id which are having the second highest salary,
under a particular manager.
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Answer / kumar mrityunjay singh
SELECT *
FROM (SELECT EMPLOYEE_ID,MANAGER_ID,SALARY,DENSE_RANK() OVER(ORDER BY SALARY DESC)RN FROM EMPLOYEES WHERE MANAGER_ID='&ENTER_MANAGER_ID')WHERE RN < '&VALUE_UPTO'
/
| Is This Answer Correct ? | 8 Yes | 4 No |
Answer / rajesh
select * from (select
emp_id,mgr_id,salary,dense_rank()over(partition by mgr_id order by salary desc nulls last) rn from emp) where rn=2;
| Is This Answer Correct ? | 5 Yes | 1 No |
Answer / chandra reddy
select empno,ename,deptno,sal from(select e.*,b.sal,rank()over(order by sal desc)send_h from em e join em1 b on e.empno=b.empno) where send_h=2
| Is This Answer Correct ? | 1 Yes | 1 No |
i have a table with the columns below as Emp_ID Address_ID Address_Line City Country -------- --------- ----------- ------ --------- Q: Display the Emp_ID's those having more than one Address_ID
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find out the second highest salary?
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