Use percentage function in pl sql

Question

Hi,
I want to get the employees that their salary are in top 13% of salaries.
I learned in this forum that I can get the 10% of top salaries in HR schema by this query:
WITH     got_tenth     AS
     SELECT     last_name, salary
     ,     NTILE (10) OVER (ORDER BY salary)     AS tenth
     FROM     hr.employees
SELECT       last_name, salary
FROM       got_tenth
WHERE       tenth     = 10
ORDER BY  salary     DESC
;Output:
LAST_NAME                     SALARY
King                           24000
Kochhar                        17000
De Haan                        17000
Russell                        14000
Partners                       13500
Hartstein                      13000
Greenberg                      12000
Errazuriz                      12000
Higgins                        12000
Ozer                           11500There is no "TOP" keyword in Oracle.I wondered, What if I want to get, for example, the employees who are in the top 12% of the salaries? here we cannot use deciles (maybe we can but it won't be very nice).
Thanks a lot to helpers

Answer 1

CUME_DIST can help:
http://docs.oracle.com/cd/B19306_01/server.102/b14200/functions035.htm
WITH  tbl  AS
  SELECT  last_name, salary
  ,  cume_dist() OVER (ORDER BY salary DESC)  AS cumulative_pcnt
  FROM  hr.employees
SELECT *
FROM tbl
WHERE cumulative_pcnt <=0.12;Tie values always evaluate to the same cumulative distribution value, so if you need exact 12% of the rows you can add employee_id in the ORDER BY list.
Edited by: 1006646 on 19.05.2013 10:14