Re: left outer join on more than 2 tables?
От | Gavin Baumanis |
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Тема | Re: left outer join on more than 2 tables? |
Дата | |
Msg-id | E8B4F516-1647-4621-B744-5DB66E1EC682@eclinic.com.au обсуждение исходный текст |
Ответ на | Re: left outer join on more than 2 tables? (Richard Broersma <richard.broersma@gmail.com>) |
Ответы |
Re: left outer join on more than 2 tables?
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Список | pgsql-sql |
At the risk of being wrong.... (I'm always ready to learn something new) - and seemingly I'm only too happy to be wrong!... And... it might even be that it is exactly the same result - but I would have proposed; SELECT R.region_name, Count(*) AS RegionComplaints FROM Region AS R LEFT JOIN City AS Ci LEFT JOIN Complaint AS Cm ON Ci.id = Cm.city_id ON R.id =C.region_id GROUP BY R.region_name; Gavin. On 17/06/2009, at 7:25 AM, Richard Broersma wrote: > On Tue, Jun 16, 2009 at 1:59 PM, Carol > Cheung<cacheung@consumercontact.com> wrote: >> I would like to find the counts of complaints by region and I would >> like all >> regions to be displayed, regardless of whether or not complaints >> exist for >> that region. Is left outer join what I'm looking for? > > SELECT R.region_name, Count(*) AS RegionComplaints > FROM Region AS R > LEFT JOIN City AS Ci > ON R.id = C.region_id > LEFT JOIN Complaint AS Cm > ON Ci.id = Cm.city_id > GROUP BY R.region_name; > > > -- > Regards, > Richard Broersma Jr. > > Visit the Los Angeles PostgreSQL Users Group (LAPUG) > http://pugs.postgresql.org/lapug > > -- > Sent via pgsql-sql mailing list (pgsql-sql@postgresql.org) > To make changes to your subscription: > http://www.postgresql.org/mailpref/pgsql-sql As always Please contact me if I can be of any further assistance. Gavin "Beau" Baumanis Senior Application Developer PalCare P/L 657 Nicholson Street Carlton North Victoria, Australia, 3054 E: beau@palcare.com.au P: +61 -3 9380 3513 M: +61 -438 545 586 W: http://www.palcare.com.au
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