Re: showing also value '0' with aggregate count()
От | wsheldah@lexmark.com |
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Тема | Re: showing also value '0' with aggregate count() |
Дата | |
Msg-id | 200109271534.LAA11176@interlock2.lexmark.com обсуждение исходный текст |
Ответ на | showing also value '0' with aggregate count() (Janning Vygen <vygen@planwerk6.de>) |
Список | pgsql-general |
You need a left outer join: select name, count(sales.client_id) from clients left outer join sales on sales.client_id = clients.id group by name; Outer joins were not supported prior to 7.1, so if you're using an earlier version, you'll want to either upgrade or read the docs for your version; they had a standard workaround prior to 7.1. Wes Sheldahl Janning Vygen <vygen%planwerk6.de@interlock.lexmark.com> on 09/27/2001 10:51:45 AM To: "PostgreSQL-General" <pgsql-general%postgresql.org@interlock.lexmark.com> cc: (bcc: Wesley Sheldahl/Lex/Lexmark) Subject: [GENERAL] showing also value '0' with aggregate count() I need some help please, i am having two simple tables: # select * from sales; client_id | product -----------+--------- 1 | toolbox 1 | nails 2 | nuts # select * from clients; id | name ----+------- 1 | peter 2 | john 3 | marc now i want to show all client name and the count of the sales in one table like this: # select name, count(sales.client_id) from clients, sales where sales.client_id = clients.id group by name; name | count -------+------- john | 1 peter | 2 works fine, but where is marc??? it should look like name | count -------+------- john | 1 peter | 2 marc | 0 who can i make it work??? i think i know why my select statement doesnt work, because of the where clause marc will never join this table because the condition will never be true... but how can i do it?? i guess its very very simple, but i just cant manage it. thanks in advance janning ---------------------------(end of broadcast)--------------------------- TIP 3: if posting/reading through Usenet, please send an appropriate subscribe-nomail command to majordomo@postgresql.org so that your message can get through to the mailing list cleanly
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