How to use group by in analytic function

Question

I need to write department which has minimum salary in one row. It must be with analytic function but i have problem with group by. I can not use min() without group by.
select * from (select min(sal) min_salary, deptno, RANK() OVER (ORDER BY sal ASC, rownum ASC) RN from emp group by deptno) WHERE RN < 20 order by deptno;
Edited by: senza on 6.11.2009 16:09

Answer 1

different query, different results.
LPALANI@l11gr2>select department_id, min(salary)
  2  from hr.employees
  3  group by department_id
  4  order by 2;
   DEPARTMENT_ID      MIN(SALARY)
              50            2,100
              20            2,100
              30            2,500
              60            4,200
              10            4,400
              80            6,100
              40            6,500
             100            6,900
                            7,000
             110            8,300
              70           10,000
              90           17,000
12 rows selected.
LPALANI@l11gr2>
LPALANI@l11gr2>-- Always lists one department in a non-deterministic way
LPALANI@l11gr2>select * from (
  2  select department_id, min(salary) min_salary
  3  from hr.employees
  4  group by department_id
  5  order by 2) where rownum = 1;
   DEPARTMENT_ID       MIN_SALARY
              20            2,100
LPALANI@l11gr2>
LPALANI@l11gr2>-- Out of the departments with the same least salary, returns the one with the least department number
LPALANI@l11gr2>SELECT   MIN (department_id) KEEP (DENSE_RANK FIRST ORDER BY salary) AS dept_with_lowest_sal, min(salary) min_salary
  2  FROM        hr.employees;
DEPT_WITH_LOWEST_SAL       MIN_SALARY
                  20            2,100
LPALANI@l11gr2>
LPALANI@l11gr2>-- This will list all the deparments with the minimum salary
LPALANI@l11gr2>select department_id, min_salary
  2  from (select
  3  department_id,
  4  min(salary) min_salary,
  5  RANK() OVER (ORDER BY min(salary) ASC) RN
  6            from hr.employees
  7            group by department_id)
  8  WHERE rn=1;
   DEPARTMENT_ID       MIN_SALARY
              20            2,100
              50            2,100