How to write this query ?

Question

how to write this query ?
list the emp name who is working for the highest avg sal department.
I can use row_number over to get correct result. If we don't use this row_number function, just plain sql, how to do it ?
the row_number version is like this
select emp.* from emp ,
select deptno, row_number() over (order by avg(sal) desc) r from emp
group by deptno
)e
where e.r = 1
and emp.deptno = e.deptno

Answer 1

Hi,
806540 wrote:
how to write this query ?
list the emp name who is working for the highest avg sal department.
I can use row_number over to get correct result. If we don't use this row_number function, just plain sql, how to do it ?ROW_NUMBER is just plain SQL, and has been since Oracle 8.1.
ROW_NUMBER (or its close relative, RANK) is the simplest and most efficient way to solve this problem. Why not do this the right way?
the row_number version is like this
select emp.* from emp ,
select deptno, row_number() over (order by avg(sal) desc) r from emp
group by deptno
)e
where e.r = 1
and emp.deptno = e.deptno
If there happens to be a tie (that is, two or more departments have the same average sal, and it is the highest), then the query above will only arbitrarily treat one of them (no telling which one) as the highest. Change ROW_NUMBER to RANK to get all departments with a claim to having the highest average sal.
You could use the ROWNUM pseudo-column instead of ROW_NUMBER, if all you want to do is avoid ROW_NUMBER.
Without using ROW_NUMBER or RANK, there are lots of ways involving other analytic functions, such as AVG and MAX.
If you really, really don't want to use analytic functions at all, you can do this:
SELECT     *
FROM     scott.emp
WHERE     deptno     IN  (
                  SELECT       deptno
                  FROM       scott.emp
                  GROUP BY  deptno
                  HAVING       AVG (sal) =  (
                                                   SELECT    MAX (AVG (sal))
                                           FROM          scott.emp
                                           GROUP BY  deptno
;