Another sql query

Question

Hi,
I have a manager table whose data is as below.
create table manager (dept_id number, manager_id number,subject varchar2(20),salary number(4,2));
insert into manager values(10, 1, subject_1,5);
insert into manager values(10, 2, subject_1,5);
insert into manager values(10, 3, subject_1,5);
insert into manager values(10, 4, subject_2,4.5);
insert into manager values(10, 5, subject_2,4.5);
insert into manager values(10, 6, subject_2,6);
insert into manager values(10, 7, subject_3,10);
insert into manager values(10, 8, subject_3,5);
insert into manager values(10, 9, 'ALL',10);
Now my requirement is to sum up the salaries on the basis of subject. If subject_1 is present then sum up salaries where subject_1 and ÁLL' and ignore other subject salaries.
If subject_2 is present then sum up salaries of all subject_2 and 'ÁLL' and ignore others.
Note that the sum of salary for each subject will always be equal that's why we are ignoring the rest if any one is present.
So the output should be like this:
dept_id salary
10 25
Thanks in advance
-D

Answer 1

Something a little more flexible perhaps...
SQL> ed
Wrote file afiedt.buf
  1  with manager as (select 10 as dept_id, 1 as manager_id, 'subject_1' as subject, 5 as salary from dual union all
  2                   select 10, 2, 'subject_1', 5 from dual union all
  3                   select 10, 3, 'subject_1', 5 from dual union all
  4                   select 10, 4, 'subject_2', 4.5 from dual union all
  5                   select 10, 5, 'subject_2', 4.5 from dual union all
  6                   select 10, 6, 'subject_2', 6 from dual union all
  7                   select 10, 7, 'subject_3', 10 from dual union all
  8                   select 10, 8, 'subject_3', 5 from dual union all
  9                   select 10, 9, 'ALL', 10 from dual
10                  )
11  -- end of test data
12  select dept_id, sum(salary) as salary
13  from (
14        select dept_id, manager_id, subject, salary
15              ,dense_rank() over (partition by dept_id order by decode(subject,'ALL',-1,to_number(replace(subject,'subject_')))) as rnk
16        from manager
17       )
18  where rnk <= 2
19* group by dept_id
SQL> /
   DEPT_ID     SALARY
        10         25
SQL>