Question

I have a file system structure with symlinks, something like this:

folder123
    index.php
    config.php -> symlink to shared/config.php
shared
    config.php

I run index.php on command line, index.php includes config.php, which is symlink to a file in some other folder.

How can I know index.php file path from the config.php script? The required result is “folder123” (well the full path of it), not “shared”. So, I cannot use __FILE__.

This works on scripts, ran by browser (at least in my case, it is the same as folder123 – the base dir):

$_SERVER['DOCUMENT_ROOT']

I need something for the command-line too.

It looks like

$_SERVER['PWD']

is something what I need, but it’s not documented, so I don’t find it a reliable way. Also, you need to combile it with SCRIPT_FILENAME to get the required results – PWD shows directory, from which the script was run, not it’s full path.

getcwd looks promising, but it can be changed by chdir – I’ll use that if I don’t find some better alternative.

Answer 1

I’m not sure if there is a way to determine what file included a file, but you can get the name of the running program (index.php in this case) with $argv[0], or possibly $_SERVER['argv'][0] if the former is not available.

//config.php
//running path
$path = dirname(realpath($argv[0]));