Executing a file only if it's included

Question

How can I code a php file to execute only if the file is included. For example in this form it won't run (or will be redirected to another page):

www.example.com/file.php

And in this form it will run:

<?php
include 'file.php';
?>

Is that possible?

http://stackoverflow.com/questions/12999658/check-if-a-file-was-included-orloaded — Henk Jansen, Dec 12 '13 at 08:23
Are you possibly looking for the solution to *store non-public files outside of the public webroot* so they can't be directly accessed by users? — deceze, Dec 12 '13 at 08:23
I didn't get the sense of "execute only if included" (if it's included as PHP file, it will be executed) - but remote files will be included as they will be processed with remote web-server. You can't get source code unless web-server treats PHP files as usual text files — Alma Do, Dec 12 '13 at 08:24

score 2 · Accepted Answer · answered Dec 12 '13 at 08:25

Add a guard expression to the start of the included file, so if you have these two files:

<?php
$runningFileName = "index.php"; 
include("included.php");
?>

<?php 
if( empty( $runningFileName ) ) die("Cannot access this page directly");
?>

merkushin · Answer 2 · 2013-12-12T08:33:34.250

2

You can define constant in main file and check it in included file.

index.php

<?php
define('IN_INDEX', true);
include __DIR__ . '/included.php';

included.php

<?php
if (!defined('IN_INDEX')) {
    redirect();
}

edited Dec 12 '13 at 08:33

answered Dec 12 '13 at 08:25

merkushin

2 Answers2