Re: Left Outer Join Question
От
Christopher Kings-Lynne
Тема
Re: Left Outer Join Question
Дата
Msg-id
GNELIHDDFBOCMGBFGEFOGEJCCBAA.chriskl@familyhealth.com.au
Ответ на
Re: Left Outer Join Question (Tom Lane)
Список
Дерево обсуждения
Re: Left Outer Join Question "Andy Marden" <amarden@usa.net>
> "Christopher Kings-Lynne" writes: > > What's probably happening is that count isn't counting NULLs? > > As per spec. If you want to count records independently of whether > any particular column is NULL or not, use count(*). See the docs, > notably > http://www.ca.postgresql.org/users-lounge/docs/7.2/postgres/sql-ex > pressions.html#SYNTAX-AGGREGATES OK, Edward - I guess Tom means do this?: select category.name, count(items.*) from category left outer join items on (category.recordnum = items.catnum) where category.recordnum = section_subcats.catnum and section_subcats.sectionnum = 1 and items.clientnum = 333 group by category.name; Chris
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