Thanks, I just posted my response to my own question for the archives. I take it also that group by is faster than distinct on. If it is a substantial performance gain I have to work on this some more. A subquery I would expect would be much of a drag, so for all keystroke-updated list-tables this would not be suitable I think.
Am 12.03.2012 um 21:57 schrieb Bartosz Dmytrak: > Hi, > You can use one of windowing function: > http://www.postgresql.org/docs/9.1/static/tutorial-window.html > http://www.postgresql.org/docs/9.1/static/functions-window.html > this could be rank() in subquery or first_value(vale any), but there could be > performance issue > > another solution could be boolean flag "default" in table address_reference > which should be unique for single company, I mean value true should be unique > - this could be reached by unique partial index on column refid_companies > with condition default = true > http://www.postgresql.org/docs/9.1/static/indexes-partial.html#INDEXES-PARTIAL-EX3 > > hope Your pg version supports windowing functions (as I remember 8.4 and > above) > > Of course there is a solution with subquery which finds min id in table > addresses of each refid_companies in table addresses_reference and this > subquery is joined with companies table, but I am afraid this is not the best > one. > > Regards, > Bartek > > > 2012/3/12 Alexander Reichstadt <l...@mac.com> > Hi, > > the following statement worked on mysql but gives me an error on postgres: > > column "addresses.address1" must appear in the GROUP BY clause or be used in > an aggregate function > > I guess I am doing something wrong. I read the web answers, but none of them > seem to meet my needs: > > SELECT > companies.id,companies.name,companies.organizationkind,addresses.address1,addresses.address2,addresses.city,addresses.zip > FROM companies JOIN addresses_reference ON > companies.id=addresses_reference.refid_companies LEFT JOIN addresses ON > addresses_reference.refid_addresses=addresses.id GROUP BY companies.id; > > > What I did now was create a view based on above statement but without > grouping. This returns a list with non-distinct values for all companies that > have more than one address, which is correct. But in some cases I only need > one address and the problem is that I cannot use distinct. > > I wanted to have some way to display a companies list that only gives me the > first stored addresses related, and disregard any further addresses. > > Is there any way to do this? > > Thanks > Alex >
