On Thu, Jul 01, 2010 at 12:37:55PM +0100, Sam Mason wrote: > On Thu, Jul 01, 2010 at 04:26:38AM -0700, David Fetter wrote: > > On Wed, Jun 30, 2010 at 12:11:35AM -0700, Nick wrote: > > > Is this the most efficient way to write this query? Id like to > > > get a list of users that have the categories 1, 2, and 3? > > > > > > SELECT user_id FROM user_categories WHERE category_id IN (1,2,3) > > > GROUP BY user_id HAVING COUNT(*) = 3 > > > > The above method depends on (user_id, category_id) being unique, > > and excludes users with, say, categories 1, 2, 3 and 4. Are you > > sure that that latter is what you want? > > AFAICT, the above code will include a user with categories 1 to 4. > Why do you think otherwise? > > If the (user_id,category_id) combination isn't unique, it's easy to > change the HAVING clause into HAVING COUNT(DISTINCT category_id) = > 3. Oops. You're right, of course. That's what I get for posting before waking up. ;) Cheers, David. -- David Fetter <david@xxxxxxxxxx> http://fetter.org/ Phone: +1 415 235 3778 AIM: dfetter666 Yahoo!: dfetter Skype: davidfetter XMPP: david.fetter@xxxxxxxxx iCal: webcal://www.tripit.com/feed/ical/people/david74/tripit.ics Remember to vote! Consider donating to Postgres: http://www.postgresql.org/about/donate -- Sent via pgsql-general mailing list (pgsql-general@xxxxxxxxxxxxxx) To make changes to your subscription: http://www.postgresql.org/mailpref/pgsql-general
