Re: mysql question

Yes...off topic... Join the table with itself. untested... SELECT t1.shopnumber, t1.item from table t1, table t2 where t1.shopnumber = 1 and t1.shopnumber = t2.shopnumber and t1.item != t2.item Something like that?? Play around with it... ---John Holmes... ----- Original Message ----- From: "scott" <scott@scottincz.com> To: <php-general@lists.php.net> Sent: Wednesday, October 30, 2002 3:43 PM Subject: [PHP] mysql question > Very slightly OT but great minds here > > I have a table with two columns and I need to get all the items that are > not in the shop number I select, with the exception of items that are in > the shop number I select. > > Table > > Shopnumber item > 1 orange > 1 banana > 1 apple > 1 pear > 2 grape > 2 coca cola > 2 pepsi > 3 orange > 4 orange > 4 pepsi > 4 7 up > 4 sunny delite > > > I need to be able to work out all the items that are in the other shops. > For example if a customer picks orange from shopnumber 1 I need to get a > result that has all items in all shops except shop1 > The problem is my query still picks items that are in shop1 if they are > in another shop as well, which is not what I need > > My current query is select * from table where shopnumber!=$shoprvar. > As an example if I use select * from table where shopnumber!=1 I would > get grape, coca-cola,pepsi (x2),7 up, sunny delite and orange (x2). I > don't want orange because it is in shop1!!!!! > > > Help and BIG THANKS for the ANSWER? > Scott > > > > > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php >