RE: New to joins - this simple one doesn't work.

2003-12-09 Thread Mike Johnson
From: fatblokeonbike [mailto:[EMAIL PROTECTED] > . > Dear List, > > My thanks to those who replied. > > I'm embarrassed as anything - the bloke I'm doing this thing > for has now changed his tiny mind and has decided to have a > set number of images per property, whereas before he was > conte

RE: New to joins - this simple one doesn't work.

2003-12-09 Thread fatblokeonbike
. Dear List, My thanks to those who replied. I'm embarrassed as anything - the bloke I'm doing this thing for has now changed his tiny mind and has decided to have a set number of images per property, whereas before he was contemplating accepting a widely variable number. I can thus stick all

RE: New to joins - this simple one doesn't work.

2003-12-09 Thread Mike Johnson
From: fatblokeonbike [mailto:[EMAIL PROTECTED] > > > $Query="SELECT properties.area, images.image_filename > > > FROM properties, images WHERE > > > properties.reference_number=images.reference_number AND > > > properties.area=$id"; > > > > > > but it doesn't work - I get the usual "...not a val

Re: New to joins - this simple one doesn't work.

2003-12-09 Thread Diana Soares
Try using '$id' : SELECT properties.area, images.image_filename FROM properties, images WHERE properties.reference_number=images.reference_number AND properties.area='$id' On Mon, 2003-12-08 at 23:18, fatblokeonbike wrote: > . > I'm new to MySQL joins with PHP and I can't get this to work - and

RE: New to joins - this simple one doesn't work.

2003-12-08 Thread Bob Loeffler
und that if you don't, and you need to make a change to a country name, it can become a big mess. Bob -Original Message- From: fatblokeonbike [mailto:[EMAIL PROTECTED] Sent: Monday, December 08, 2003 4:39 PM To: Mike Johnson Cc: [EMAIL PROTECTED] Subject: RE: New to joins - this sim

RE: New to joins - this simple one doesn't work.

2003-12-08 Thread fatblokeonbike
. > $Query="SELECT properties.area, images.image_filename FROM > properties, > images WHERE properties.reference_number=images.reference_number AND > properties.area=$id"; > > but it doesn't work - I get the usual "...not a valid MySQL > result resource" > If you can help, thanks in advance. If $i

RE: New to joins - this simple one doesn't work.

2003-12-08 Thread Mike Johnson
From: fatblokeonbike [mailto:[EMAIL PROTECTED] > $Query="SELECT properties.area, images.image_filename FROM > properties, > images WHERE properties.reference_number=images.reference_number AND > properties.area=$id"; > > but it doesn't work - I get the usual "...not a valid MySQL > result res