Use Variable in Included?

D_tunisia

I am using include to add a piece of code to a page and I want to choose which code via a query in the URL string.

The code I am using is:

include("functions/{$page}");

For example, if a user goes to http://www.mysite.com/index.php?code=register.php then the code included would be from register.php.

But I am getting this error with the above code:

Warning: main(functions/{$page}): failed to open stream: No such file or directory in /home/jclickth/public_html/page_test.php on line 0

Warning: main(): Failed opening 'functions/{$page}' for inclusion (include_path='.:/usr/lib/php:/usr/local/lib/php') in /home/jclickth/public_html/page_test.php on line 0

Can anyone help me out?

thorpe

Your link would need to be...

http://www.mysite.com/index.php?page=register.php

And your code...

include("functions/{$_GET['page']}");

Be aware that there are alot of security issues with doing things this way. Validate the include first. Otherwise.... anyone could run php on your site.

Roger_Ramjet

Use a switch

url =
http://www.mysite.com/index.php?foo=1

switch ($_GET['foo']) {
case 1:
   $uri = 'register.php';
case 2:
   $uri = 'login.php';
case 3:
   $uri = 'main.php';
default:
   $uri = "index.php';
}

include ("$uri");