Re: [PHP] always last

2001-08-24 Thread Chris Hayes
Van:"Andrey Hristov" <[EMAIL PROTECTED]> Aan:<[EMAIL PROTECTED]> Datum verz. Thu, 23 Aug 2001 18:33:53 +0300 Onderwerp: Re: [PHP] always last > Use a counter when fetching from the DB > if you use whi

Re: [PHP] always last

2001-08-24 Thread Chris Hayes
;[EMAIL PROTECTED]> Aan: <[EMAIL PROTECTED]> Datum verz. Thu, 23 Aug 2001 11:10:25 -0400 Onderwerp: [PHP] always last > Hi, I need a little bit of help. > My variable $team in the option block is always set to the last variable that > i

Re: [PHP] always last

2001-08-23 Thread Dave Freeman
On 23 Aug 01, at 11:10, Jeremy Morano wrote: > $result = @mysql_query($sql,$connection) or die("Couldn't execute query."); > > > while ($row = mysql_fetch_array($result)) { > $uid = $row['uid'] > $team = $row['team']; if ($uid == $row['uid']) { $option_block .= "$team"; } else { >

Re: [PHP] always last

2001-08-23 Thread Andrey Hristov
lt;[EMAIL PROTECTED]> To: <[EMAIL PROTECTED]> Sent: Thursday, August 23, 2001 6:10 PM Subject: [PHP] always last > Hi, I need a little bit of help. > My variable $team in the option block is always set to the last variable > that is readI would like it to contain t

[PHP] always last

2001-08-23 Thread Jeremy Morano
Hi, I need a little bit of help. My variable $team in the option block is always set to the last variable that is readI would like it to contain the option that the user clicks.Please help me! My Select works properly ... $result = @mysql_query($sql,$connection) or die("Couldn't exe