passing 'include();' as a variable

[pahikua]

I'm trying to pass an include statement through a variable but I'm not having any luck.

Here is an example:

 <? $var = "include('somefile.php')"; ?>

and then later on:

<? $var ?>

to actually include the 'somefile.php'

I tried:

<? echo $var ?>

but that just prints out the include statement as a string.

It works if I hard code the include into the script and just place the file name in a path like this:

$var = "somefile.php";

include("$somefile.php");

but then I can only include a file and I need the option to either include a file or just print out html directly.

Any suggestions?

Thanks for the help.

Pahikua

[eoghain]

Use eval() http://www.php.net/manual/en/function.eval.php:

<?php
   $var = "include('somefile.php');";
   eval($var);
?>

[pahikua]

Using eval() worked. thanks