php.net |  support |  documentation |  report a bug |  advanced search |  search howto |  statistics |  random bug |  login
Bug #4013 header function not functions after and if statment
Submitted: 2000-04-03 14:13 UTC Modified: 2005-03-31 16:13 UTC
From: everett at bentonrea dot com Assigned:
Status: Wont fix Package: Misbehaving function
PHP Version: 3.0.14 OS: linux 2.2.5-15smp
Private report: No CVE-ID: None
Welcome back! If you're the original bug submitter, here's where you can edit the bug or add additional notes.
If you forgot your password, you can retrieve your password here.
Password:
Status:
Package:
Bug Type:
Summary:
From: everett at bentonrea dot com
New email:
PHP Version: OS:

 

 [2000-04-03 14:13 UTC] everett at bentonrea dot com
Can some one explain this to me?


     $admin_test = admin_user($REMOTE_USER);
     if( !$admin_test ) { header("Location: index.phtml"); exit; }
     exit();

This is the problem portion of my code.

If I run this code, and the $admin_test is false I get the "Warning Can't display more header info....."  My initial thought was \n or \r in the function, but I have even changed the function to a simple return false; line.

If $admin_test is true, then the page exits, and there is NOTHING that shows up when I do the View Source.

If I remove the if statement, and just run the header function, like so:

     $admin_test = admin_user($REMOTE_USER);
     header("Location: index.phtml");
     exit();

It works fine and redirects to that correct page.

Any one have any ideas.

Apache 1.3.11
PHP 3.0.14
mod_perl 1.21
mod_ssl.5.0
mod_auth_mysql 2.20
mysql 3.22.32
gd 1.7.3

Patches

Pull Requests

History

AllCommentsChangesGit/SVN commitsRelated reports
 [2005-03-31 16:13 UTC] php-bugs at lists dot php dot net
We are sorry, but we do not support PHP 3 related problems anymore.
 
PHP Copyright © 2001-2025 The PHP Group
All rights reserved.
Last updated: Tue May 13 01:01:28 2025 UTC