Answer / rahul tripathi

The Genral answer which can work on any SQL:-

select salary from emp e1 where (n-1)=(select count(*)
from emp where salary > e1.salary )

Rahul Tripathi

rahultripathi@inkanpur.com

Answer / brahma

select * from table_name t1 where (n-1)=
(select count(distinct(column_name)) from table_name t2
where t2.column_name>t1.column_name)

Answer / kavitha.r

Sorry. This one is correct


select top 1 salary from employee where salary in (select
distinct
top n salary from employee order by salary desc) order by
salary asc

Answer / udathayadav

select * from tablename.aliasname where nth=select
count(sal) from tablename where aliasname.sal<=sal

Answer / nidhish

replace the N with wat ever number u want

select min(salary) from employee where salary in ( select
distinct top N salary from employee order by salary desc)

Answer / debasish

select salary from table t where n-1=(select distinct count
(sal) from table t1 where t1.sal>t.sal)

Answer / amit

select * from emp as e1 where sal (n-1) = select distinct
(count(*) from emp as e2 where e2.sal>e1.sal)

Where n is no that u want which postion's salary u want
suppose second higest means n=2

Amit

Answer / ayush sharma

select sal from (select distinct sal from emp order by desc)
where rownum <=n minus select sal from(select distinct sal
from emp order by desc) where rownum<=n-1;

Answer / ramakrishna

select e1.* from emp e1 where n=(select
count(distinct(e2.sal)) from emp e2 where e1.sal<=e2.sal);

Answer / bharath

select distinct(sal) from emp a where 3 = (select
count(distinct(sal)) from emp b where a.sal<= b.sal)

replace 3 by 2 for second highest...

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