PHP script detect that it is an include

Question

I have a PHP class file that is used as an include both through a web server and by a cron process. I am looking for a way to add some code to the head of the script so that I can detect if the script is being launched directly from the command line instead of being included in another script. This way I can make testing a bit easier by calling the script directly and having a function instantiate the object and execute some code on it without needing to create a wrapper script for it.

I tried using if (php_sapi_name() == 'cli') { but that tests true even if the script is being included in another script that was called from the command line.

This is a duplicate. See stackoverflow.com/questions/2397004/… for an answer. There are a few valid ones. — Ryan Matthews
– Ryan Matthews, Commented Apr 20, 2011 at 15:33
Isn't that more about detecting (and preventing) direct access through Apache, rather than what I want to do, detecting and performing additional operations when the file is directly accessed at the command line? — Wige
– Wige, Commented Apr 20, 2011 at 15:47
exactly. Your question is definitely not a duplicate of that one. — Capsule
– Capsule, Commented Apr 20, 2011 at 15:49

Alexander Garden · Accepted Answer · 2012-05-27 01:09:41Z

5

Checking that $argv[0] is set lets you know if things were invoked from the command line; it does not tell you if the file was invoked directly or if it was included from another file. To determine that, test if realpath($argv[0]) == realpath(__FILE__).

Putting it all together:

if (isset($argv[0]) && realpath($argv[0]) == realpath(__FILE__))
    // This file was called directly from command line

edited May 27, 2012 at 1:09

answered May 27, 2012 at 0:57

Alexander Garden

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1 Comment

Sdlion Over a year ago

This is the real answer. $argv[0] will be set on both the file that's being called through the cli and scripts that will be included by that script. But on all scripts $argv[0] will be set with the same filename, the one that's being directly called. __FILE__ != $argv[0] on included/required scripts.

jeroen · Accepted Answer · 2011-04-20 15:39:30Z

3

You can check for the presence of $argv. If $argv[0] is set (the script name), your script is run from the command line.

answered Apr 20, 2011 at 15:39

jeroen

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4 Comments

Wige Over a year ago

Won't this still test true if the script is included by another script that was called from the command line? Although I could check the value of $argv[0] if it is set... Hm...

jeroen Over a year ago

@Wige It´s a global variable so if your script is called from another script that is run from the command line, it will still exist.

Wige Over a year ago

This looks like the closest solution so far. What seems to work at the moment is if (isset($argv[0]) && strpos($argv[0], 'file.class.php') > 0) { This method seems to be the best to always detect if the file is being called directly from the command line.

jeroen Over a year ago

@Wige Yes, you´ll have to add a second if the script that includes it can also be run from the command line.

Capsule · Accepted Answer · 2011-04-20 15:53:04Z

You could test if $_SERVER['DOCUMENT_ROOT'] is empty. It will be empty in the command line execution since it's a variable predefined by apache.

What's interesting is $_SERVER['DOCUMENT_ROOT'] is defined but empty. I would have guessed it wouldn't be defined at all when using the cli.

You could also test if $argv is defined or its size (at least 1 when using CLI). I didn't test when including the file but if defined, sizeof($argv)would definitely be 0.

Other possible tests are $_SERVER['argc'] (0 when executed by a server, 1 when executed from CLI) and a strange $_SERVER['_'] defined to the path to the PHP binary, which is not defined at all when served.

To conclude, I would rely on $_SERVER['argc'] (or $argc) which is a direct count of the number of arguments passed to the command line: will always be 0 when the script is served. Relying on $argv is more complicated: you have to test if $argv[0] is set or if sizeof($argv) is > 0.

eaj · Accepted Answer · 2011-04-20 15:42:07Z

0

I have successfully used if (defined('STDIN')) as such a check. STDIN will be defined if the script is run from a shell, but not if it's in a server environment.

As you can see here and at the link Ryan provided in his comment, there are lots of possibilities.

answered Apr 20, 2011 at 15:42

eaj

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Comments

Alvy · Accepted Answer · 2022-04-05 08:33:35Z

0

if($_SERVER['SCRIPT_FILENAME']==__FILE__){
   // Directly
}else{
   // Included
}

answered Apr 5, 2022 at 8:33

Alvy

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1 Comment

Andrea Olivato Over a year ago

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