HR Schema: Select salary which is earned by the most employees+how often

Question

hi all,
select * from (
SELECT salary, COUNT(salary)
FROM employees
group by salary order by 2 desc) where rownum < 2;
SALARY | COUNT(SALARY)
2500 | 6
Is there another solution for this result (without rownum)?
thanks

Answer 1

user13428495 wrote:
thanks a lotYour solution was way better then all others posted so far. Why? Because you only scanned the employees table one time. All others needed to scan this table two times. Therefore your query will probably execute faster if the data will become more. For small data sets you will not notice any difference.
Other solutions that will only scan the table one time can be using the KEEP syntax or an analytic function.
untested
/* aggregate version */
select max(count_sal), max(salary) keep (dense_rank first order by count_emps desc)
from (SELECT salary, COUNT(*) as count_emps
        FROM employees
       group by salary )
/* maybe the double group works directly without extra select */
select max(COUNT(*)) cnt, max(salary) keep (dense_rank first order by COUNT(*) desc) salary
FROM employees
group by salary;
/* analytic version with a different output*/
select v2.* from (
   select v1.*, rank() over (order by count_emps desc)
   from (
      SELECT e.*, count(*) over (partition by e.salary) count_emps rn
      FROM employees e) v1
   order by count_emps desc) v2
where v2.rn = 1Edited by: Sven W. on Sep 8, 2011 1:53 PM typo correction: dense_rank