[RESOLVED] Include a URL with a variable in it

ajd344

Hi! I need to use the following code:

<?
include "gbook.php?page=2";
?>

When I run it, I get this:

Warning: include(gbook.php?page=2) [function.include]: failed to open stream: Invalid argument in D:\XAMPP\xampp\htdocs\ATown\test.php on line 2

Warning: include() [function.include]: Failed opening 'gbook.php?page=2' for inclusion (include_path='.;D:\XAMPP\xampp\php\pear\') in D:\XAMPP\xampp\htdocs\ATown\test.php on line 2

Any ideas on how to include this? Thanks!

Rodney_H_

If you are including the file, you won't need to pass the variable via the query string.

If the variable has an assigned value, it will "pass" that value into the included file, because the included file will "become" part of the script that runs it.

for instance, if you have:

<?php
// assign value to var
$page = 2;

// include file
include_once('gbook.php');

// etc...
?>

the "gbook.php" file will be embedded into the script the runs it and the value of $page will be accessible...

Hope that helps.

ajd344

Oh! I get it! Thanks a lot for your help, it works now!

Houdini

I noticed that you are using short tags i.e.

<?
include "gbook.php?page=2";
?>

I know that this code is running on your local machine due to the address of drive D and the XAMPP. You should try to not use short tags because if the host you are using for live internet scripts upgrades to a newer PHP version it is entirely possible that short tags will be turned off, and when this happens you scripts will not work, however if you make a habit of always using

<?php//this always works <? does not
include "gbook.php";//fixed the error here
?>