Showing posts with label zsource. Show all posts
Showing posts with label zsource. Show all posts

Tuesday, February 14, 2012

counting distincts

This won't work...
SELECT count(distinct z.id, c.alias, c.type)
FROM customer c JOIN zsource z ON c.id=z.id
How can I count the number of distinct combinations of 3 separate fields?This should work :
SELECT count(*) FROM (
SELECT distinct z.id, c.alias, c.type
FROM customer c JOIN zsource z ON c.id=z.id) a
-Nags
"Dave" <dave@.nodomain.tv> wrote in message
news:#H8rx9bFEHA.2576@.TK2MSFTNGP11.phx.gbl...
> This won't work...
> SELECT count(distinct z.id, c.alias, c.type)
> FROM customer c JOIN zsource z ON c.id=z.id
> How can I count the number of distinct combinations of 3 separate fields?
>|||This is one way. There could be other better ways.
select count(RecCount)
from (select count(*) as RecCount
from customer c JOIN zsource z ON c.id=z.id
group by z.id, c.alias, c.type) d
"Dave" <dave@.nodomain.tv> wrote in message
news:%23H8rx9bFEHA.2576@.TK2MSFTNGP11.phx.gbl...
> This won't work...
> SELECT count(distinct z.id, c.alias, c.type)
> FROM customer c JOIN zsource z ON c.id=z.id
> How can I count the number of distinct combinations of 3 separate fields?
>|||Thanks much
"Nags" <nags@.DontSpamMe.com> wrote in message
news:O4lRaPcFEHA.2560@.TK2MSFTNGP12.phx.gbl...
> This should work :
> SELECT count(*) FROM (
> SELECT distinct z.id, c.alias, c.type
> FROM customer c JOIN zsource z ON c.id=z.id) a
> -Nags
> "Dave" <dave@.nodomain.tv> wrote in message
> news:#H8rx9bFEHA.2576@.TK2MSFTNGP11.phx.gbl...
fields?
>