Reputation: 194
I have this table (I put the name over needed colums)
iddip date idv idc val
47 2018-06-01 00:00:00.000 0 3 3 60 NULL NULL
47 2018-06-01 00:00:00.000 0 1 3 200 NULL NULL
47 2018-06-01 00:00:00.000 0 1 4 280 NULL NULL
43 2018-06-01 00:00:00.000 0 3 2 510 NULL NULL
53 2018-06-01 00:00:00.000 0 1 4 480 NULL NULL
29 2018-06-01 00:00:00.000 0 3 2 510 NULL NULL
2 2018-06-11 00:00:00.000 0 1 2 480 NULL NULL
47 2018-06-02 00:00:00.000 0 1 3 100 NULL NULL
I want to obtain this:
id idc Totidv1 Totidv3 TOT
47 3 300 60 360
47 4 280 0 280
43 2 0 510 510
53 4 480 0 480
29 2 0 510 510
2 2 480 0 480
The closest I can get is:
SELECT DISTINCT(iddip),IDCENTROCOSTO,tot=SUM(VALORE),ord=( SELECT SUM(isnull(VALORE,0)) FROM VALORIVOCICDC WHERE IDVOCE='1' and iddip=v.IDDIP and IDCENTROCOSTO ='3' GROUP BY iddip,IDCENTROCOSTO),
str=( SELECT SUM(isnull(VALORE,0)) FROM VALORIVOCICDC WHERE IDVOCE='3' and iddip=v.IDDIP and IDCENTROCOSTO ='3' GROUP BY iddip,IDCENTROCOSTO)
FROM VALORIVOCICDC v
GROUP BY v.iddip,IDCENTROCOSTO
But it returns wrong sums in totidv1 and totisv3, How can I do this? Thanks for any hint
Upvotes: 0
Views: 30
Reputation: 50200
You just need a GROUP BY here (not distinct) and a couple of CASE statements:
SELECT
id,
idc,
SUM(CASE WHEN idv=3 THEN idv ELSE 0 END) as totidv1,
SUM(CASE WHEN idv=1 THEN idv ELSE 0 END) as totidv3,
SUM(idv) as Tot
FROM yourtable
GROUP BY id, idc
Note that Distinct
is not a function that you can call like SELECT DISTINCT(somecolumn)
This is functionally equivalent to SELECT DISTINCT somecolumn...
in that it works against the entire record set returned by the SELECT statement either way.
Upvotes: 1