Re: [PHP-DB] Brain not working, help needed :-)

2005-08-29 Thread Jordan Miller
you can also do this to save room without having to create $varname: ${"id".$num} = "some text"; Jordan On Aug 29, 2005, at 12:10 AM, Micah Stevens wrote: Hi Chris, You can use variable variables.. like this: $num = 2 $varname = "id".$num; $$varname = "id2 is stored here!"; It's in the

RE: [PHP-DB] Brain not working, help needed :-)

2005-08-29 Thread ziv gabel
The format should be $id[$count] -Original Message- From: Chris Payne [mailto:[EMAIL PROTECTED] Sent: Monday, August 29, 2005 8:46 AM To: php-db@lists.php.net Subject: [PHP-DB] Brain not working, help needed :-) Hi there everyone, I have the following code: while ($row

RE: [PHP-DB] Brain not working, help needed :-). .

2005-08-29 Thread Norland, Martin
Figured I'd expand on what these evaluate to, in case anyone gets lost otherwise: >-Original Message- >From: Micah Stevens [mailto:[EMAIL PROTECTED] [snip] >You can use variable variables.. like this: > >$num = 2 $num == 2 [obviously :) ] >$varname = "id".$num; $varname == "id2" >$$varn

Re: [PHP-DB] Brain not working, help needed :-)

2005-08-28 Thread Micah Stevens
Hi Chris, You can use variable variables.. like this: $num = 2 $varname = "id".$num; $$varname = "id2 is stored here!"; It's in the docs in the variables section.. HTH, -Micah On Sunday 28 August 2005 11:45 pm, Chris Payne wrote: > Hi there everyone, > > > > I have the following code: > > > >

[PHP-DB] Brain not working, help needed :-)

2005-08-28 Thread Chris Payne
Hi there everyone, I have the following code: while ($row = mysql_fetch_array($sql_result)) { $id = $row["id"]; $video_link = $row["video_link"]; $video_thumbnail = $row["video_thumbname"]; $video_title = $row["video_title"]; $video_description = $row["vide