Re: Count equals 0
От | greg@turnstep.com |
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Тема | Re: Count equals 0 |
Дата | |
Msg-id | ea0824a43c82b86591e6b913514cf5ba@biglumber.com обсуждение исходный текст |
Ответ на | Count equals 0 (Rudi Starcevic <rudi@oasis.net.au>) |
Список | pgsql-sql |
-----BEGIN PGP SIGNED MESSAGE----- Hash: SHA1 > I've tries Left outer join's with and without coalesce without joy so far. > Please check out my query. Your query is good, let's just make the whole thing a subselect and use an outer join to get the missing sites: SELECT u2.site_name, COALESCE(u1.total, 0) FROM ( SELECT site_name, COUNT(cus_id) AS total FROM user_main um WHERE NOT EXISTS ( SELECT cus_id FROM user_sys_messages usm WHERE usm.cus_id = um.cus_id ) GROUP BY site_name ) AS u1 RIGHT OUTER JOIN user_main u2 USING (site_name) GROUP BY 1,2 ORDER BY u2.site_name; You could also write this as a LEFT OUTER JOIN (by putting the subselect second), but I think it is clearer if the USING or ON is as close as possible to the JOIN statement. - -- Greg Sabino Mullane greg@turnstep.com PGP Key: 0x14964AC8 200303181009 -----BEGIN PGP SIGNATURE----- Comment: http://www.turnstep.com/pgp.html iD8DBQE+dzbwvJuQZxSWSsgRAj5FAJ4s2ZJgaBm8g4otdHHNI6VkhoElsgCg0Bkb drxn19A1moxzQbAFy3Jv+Dc= =tImk -----END PGP SIGNATURE-----
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