How do I join a select distinct to a select sum?

Question

I am trying to join a select disinct statement to a select sum statement. Here is an example of the data

CD  STDATE    ENDDATE   PR
F1  01/02/09  01/04/09  $10
F1  01/02/09  01/04/09  $40
F1  01/02/09  01/04/09  $20
F1  01/02/09  01/04/09  $30
F1  01/22/09  01/26/09  $10
F1  01/22/09  01/26/09  $50
F1  01/22/09  01/26/09  $20

My desired output is -

F1  01/02/09  01/04/09  $100
F1  01/22/09  01/26/09  $80

So far, I have the foll statement, which is incorrect:

SELECT DISTINCT STDATE, ENDDATE, CD FROM TABLE AS A 
INNER JOIN (SELECT SUM(PR) FROM TABLE) AS B 
ON A.STDATE=B.STDATE

How can I do that???

Answer 1

SELECT CD,STDATE,ENDDATE,SUM(PR) FROM TABLE GROUP BY CD,STDATE,ENDDATE

(missed CD in original answer)

Answer 2

Select cd, stdate, enddate, sum(PR)
from Table
group by cd, stdate, enddate

Answer 3

I think you are looking for grouping:

select
    t.cd,
    t.stdate,
    t.enddate,
    sum(t.pr)
from
    table t
group by
    t.cd,
    t.stdate,
    t.enddate

Answer 4

simple grouping and summing, pick up any SQL programming book

SELECT CD, STDATE,ENDDATE,SUM(PR) as SomeSum
FROM TABLE
GROUP BY CD, STDATE,ENDDATE