SQL Aggregate Function

Question

I am trying to create a query that will output a single row for each employee. Unfortunately, I can't figure out how to group by the aggregate function itself. Is there a way to do this in SQL Developer? Here is my code at this point:
COL ename FORMAT A25 HEADING Employee|Name
COL salary FORMAT $999,999.99 HEADING Salary
COL comm FORMAT A10 HEADING Commission|Rate
COL earn FORMAT $999,999,999.99 HEADING Sept/2010|Earnings
SELECT DISTINCT e.last_name||', '||e.first_name "ename",salary,
CASE WHEN commission_pct > 0 THEN
commission_pct
ELSE 0
END AS comm,
CASE WHEN commission_pct > 0
AND o.total > 0 THEN
(SUM(o.total)*(commission_pct*.01)+salary)
ELSE salary
END AS earn
FROM (SELECT sales_rep_id,total
FROM s_ord
WHERE date_ordered LIKE '%SEP-10') o
RIGHT OUTER JOIN s_emp e
ON e.id = o.sales_rep_id
GROUP BY e.last_name,e.first_name,e.salary,e.commission_pct,o.total;

Answer 1

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SELECT DISTINCT e.last_name
                ||', '
                ||e.first_name "ename",
                salary,
                CASE
                  WHEN commission_pct > 0 THEN commission_pct
                  ELSE 0
                END            AS comm,
                CASE
                  WHEN commission_pct > 0
                       AND o.total > 0 THEN ( SUM(o.total) * (
                                              commission_pct * .01 ) + salary )
                  ELSE salary
                END            AS earn
FROM   (SELECT sales_rep_id,
               total
        FROM   s_ord
        WHERE  date_ordered LIKE '%SEP-10') o
       right outer join s_emp e
                     ON e.id = o.sales_rep_id
GROUP  BY e.last_name,
          e.first_name,
          e.salary,
          e.commission_pct,
          o.total;