Re: [PHP] giving variable value in URL problem

2003-11-16 Thread Burhan Khalid
Chris Shiflett wrote: --- Burak Delice <[EMAIL PROTECTED]> wrote: http://localhost/menu.php?status=0 php return [snip] if ($status==0) echo '"trying"; if (!$_GET['status']) { echo '"trying"; } Firstly, !$_GET['status'] means "if status is not equal to or greater than zero" -- which of

Re: [PHP] giving variable value in URL problem

2003-11-16 Thread Jason Wong
On Sunday 16 November 2003 13:31, Burak Delice wrote: Please do not top-post, it makes replying in a coherent fashion *much* harder. I've corrected it for you. > "Chris Shiflett" <[EMAIL PROTECTED]> wrote in message > news:[EMAIL PROTECTED] > > > --- Burak Delice <[EMAIL PROTECTED]> wrote: > > >

Re: [PHP] giving variable value in URL problem

2003-11-15 Thread Burak Delice
thanx Chris, this code is running on Web, is not on Local. and I did your way, it did not work regards "Chris Shiflett" <[EMAIL PROTECTED]> wrote in message news:[EMAIL PROTECTED] > --- Burak Delice <[EMAIL PROTECTED]> wrote: > > http://localhost/menu.php?status=0 php return > > [snip] > > > if

Re: [PHP] giving variable value in URL problem

2003-11-15 Thread Chris Shiflett
--- Burak Delice <[EMAIL PROTECTED]> wrote: > http://localhost/menu.php?status=0 php return [snip] > if ($status==0) echo '"trying"; if (!$_GET['status']) { echo '"trying"; } Hope that helps. Chris = Chris Shiflett - http://shiflett.org/ PHP Security Handbook Coming mid-2004 HT