how to express DENSE_RANK with ANSI SQL?

Question

Is it possible to express the statment below by ANSI SQL? Since the example below belongs to PL/SQL. Thanks in advance.

SELECT department_id,
MIN(salary) KEEP (DENSE_RANK FIRST ORDER BY commission_pct) "Worst",
MAX(salary) KEEP (DENSE_RANK LAST ORDER BY commission_pct) "Best"
   FROM employees
   GROUP BY department_id;

DEPARTMENT_ID      Worst       Best
------------- ---------- ----------
           10       4400       4400
           20       6000      13000
           30       2500      11000
           40       6500       6500
           50       2100       8200
           60       4200       9000
           70      10000      10000
           80       6100      14000
           90      17000      24000
          100       6900      12000
          110       8300      12000
                    7000       7000

Answer 1

This returns the same result (as far as I can tell) but does not need a join and is ANSI SQL:

select department_id,
       min(case when min_comm = 1 then salary end) as worst,
       max(case when max_comm = 1 then salary end) as best
from (
  select department_id, 
         salary, 
         dense_rank() over (partition by department_id order by commission_pct desc) as max_comm,
         dense_rank() over (partition by department_id order by commission_pct) as min_comm
  from employees
) t
group  by department_id
order by 1;

Answer 2

You can use self-joins instead of analytics in most cases. Here's an equivalent of your MAX(salary) KEEP (DENSE_RANK LAST ORDER BY commission_pct):

SELECT department_id, MAX(salary)
  FROM employees e
 WHERE (department_id, commission_pct) IN
       (SELECT department_id, 
               MAX(commission_pct) 
          FROM employees 
         GROUP BY department_id)
 GROUP BY department_id

Obviously getting both the MAX and MIN would be a bit trickier (and uglier) but is doable.

Answer 3

Having given this some more thought, I'll take a stab at it. I think you are trying to show the highest and lowest value (salary in the example) for each group(deptid) using ANSI SQL? Here's a super-simplistic example of it (using DENSE_RANK()) that should work on any ANSI compliant db:

SQL FIddle

select
    t1.deptid,
    t1.salary as Highest,
    t2.salary as Lowest
from
    (
    select
        deptid,
        salary,
        dense_rank() over (partition by deptid order by salary desc) as First  --rank desc for highest
    from
        salaries
    ) T1
inner join 
    (
    select
        deptid,
        salary,
        dense_rank() over (partition by deptid order by salary asc) as worst -- rank ascending for lowest
    from
        salaries
    )T2
on 
    t1.deptid = t2.deptid

where
    t1.first = 1  --GET THE HIGHEST SALARY
    and t2.worst = 1 -- GET THE LOWEST

Using min and max would also work and be a lot simpler:

Another SQL Fiddle

select
  deptid,
  max(salary) as Highest,
  min(salary) as Lowest
from
  salaries
group by
  deptid

Answer 4

This construct avoids additional join to employees table. In ANSI SQL you would have to 1st query the highest commission_pct per department and then join the employees (again) to find salaries which have this highest commission_pct

In ANSI SQL it would like something like this:

select * from
(
  SELECT department_id,
   MIN(commission_pct) max_c
   MAX(commission_pct) min_c
  FROM employees
  GROUP BY department_id
) e1 
join employees e2 
  on (e1.department_id = e2.department_id and e1.max_c = e2.commission_pct)
join employees e3 
  on (e1.department_id = e3.department_id and e1.min_c = e3.commission_pct)

Even this is not 100% correct.