Answer / sujit pingale

Ans: select distinct(a.salary) from customers a where &n =
(select count(distinct(b.salary)) from customers b where
a.salary <= b.salary);

It will ask you to enter number in input box which salary
you want to see

Answer / sujitpingale

Ans: select distinct(a.salary) from customers a where &n =
(select count(distinct(b.salary)) from customers b where
a.salary <= b.salary);

Enter value in inputbox. If you enter 1 then it will give
highest salary

Answer / ajit

select e.*, rn
from ( select empno, ename, sal, deptno, dense_rank() over ( order by sal desc ) rn
from emp) e
where rn = & rn;

Answer / abhishekjaiswal

SELECT A.FIRST_NAME,
       A.SALARY
   FROM EMPLOYEES A
   WHERE 3 = ( SELECT COUNT(*) -- Replace 3 with any value of (N - 1)
               FROM EMPLOYEES B
               WHERE B.SALARY > A.SALARY)

Answer / pavan

select e.* from emp e where &n=(select count(distinct e1.sal)from emp e1 where e.sal>e1.sal)

Answer / ajit

select *
from emp
where sal = ( select max(sal)
from emp
where level = &n
connect by prior sal > sal
group by level);

Can we create index on primary key?

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