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This Error On The Site Do Not Know O'que Can Be


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Posted

You probably use $_GET to get something from the empty URL.

http://www.mediafire.com/download/c96fwa2x2an1pfm/Web%282%29.rar

 

u could of a look. because I do not understand very well site What do I could do to try to fix this error?

 

I found this php here's a look

<?php
switch($_GET['pag']){
	default:
	include "home.php";
	break;
	
	case 'info':
	include "info.php";
	break;
	
	case 'cp_registro':
	include "cp_registro.php";
	break;	

	case 'cp_mudar':
	include "cp_mudar.php";
	break;	
	
	case 'cp_recuperar':
	include "cp_recuperar.php";
	break;	
	
	case 'downloads':
	include "downloads.php";
	break;	

	case 'doacoes':
	include "doacoes.php";
	break;	
	
	case 'e_geral':
	include "e_geral.php";
	break;	
	
	case 'e_clans':
	include "e_clans.php";
	break;	

	case 'e_sieges':
	include "e_sieges.php";
	break;	

	case 'doacoes_confirmar':
	include "doacoes_confirmar.php";
	break;	

	case 'donate':
	include "donate.php";
	break;
	
	case 'chat':
	include "chat.php";
	break;

}
?>
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Posted (edited)

Check if isset($_GET['pag']), so if it is - get to the switch, else include the default page.

Edited by Tessa
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Posted

try my metod :)

<?php
$script = basename( $_GET['page'] );
$dir = 'pages/';
$file = $dir . $script . '.php';
if ( !file_exists( $file ) ) {
  $file = $dir . 'default.php';
}
require( $file );
?>

link will be index.php?page=contacts

  • 0
Posted (edited)

try my metod :)

 

<?php$script = basename( $_GET['page'] );$dir = 'pages/';$file = $dir . $script . '.php';if ( !file_exists( $file ) ) {  $file = $dir . 'default.php';}require( $file );?>
link will be index.php?page=contacts
You will still get this notice if $_GET is empty. So, use isset() which checks if 'page' is in the array.

if (isset($_GET['page'])) {
    // Code depending on $_GET here...
}
I would suggest Symfony's HttpFoundation which takes care about the request/response. Edited by Tessa

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