Question

I have a file that calls a set of functions utils.php, however when I include it, my show_source function includes the source of the included file. I tried putting the function file in a parent folder, but even calling..

include("../includes/utils.php");

Still shows the source of the file. How do I disable the source for any included files? Thanks in advanced!

Edit: I found out why it was showing the source, it was some weird PHP issue my school has.. sorry for the confusion as well, but showing the source was a parameter of the assignment

Answer 1

If you want to show_source (for whatever reason, in this case it’s required) but leave out the source of included files, BUT you have to show the source of the included file if it is linked to…

<?php
function test($param) {
    //do something
}

if ($_GET['show_source'] == "true") {
    show_source(__FILE__);
}

and to link to it..

<a href='../includes/utils.php?show_source=true'>Show Source of Utils.php</a>

I’m not sure if this helps everybody, but it’s how I solved this issue.