Check if a file was included or loaded

Question

Is there any elegant way to check if a file was included by using include/include_once/require/require_once or if the pag

Answer 1

They is no way to separate them as include/include_once/require/require_once but php has get_included_files and get_required_files which is the same thing and only returns array of all included files. Its does not separate it if its required or included.

Example a.php

include 'b.php';
include_once 'c.php';
require 'd.php';
var_dump(get_required_files());

Output

array
  0 => string '..\lab\stockoverflow\a.php' (length=46) <---- Returns current file
  1 => string '..\lab\stockoverflow\b.php' (length=46)
  2 => string '..\lab\stockoverflow\c.php' (length=46)
  3 => string '..\lab\stockoverflow\d.php' (length=46)

But you can do something like

$inc = new IncludeManager($file);
var_dump($inc->find("b.php")); // Check if a file is included
var_dump($inc->getFiles("require_once")); // Get All  Required Once 

Class Used

class IncludeManager {
    private $list = array();
    private $tokens = array();
    private $find;
    private $file;
    private $type = array(262 => "include",261 => "include_once",259 => "reguire",258 => "require_once");

    function __construct($file) {
        $this->file = $file;
        $this->_parse();
    }

    private function _parse() {
        $tokens = token_get_all(file_get_contents($this->file));
        for($i = 0; $i < count($tokens); $i ++) {
            if (count($tokens[$i]) == 3) {
                if (array_key_exists($tokens[$i][0], $this->type)) {
                    $f = $tokens[$i + 1][0] == 371 ? $tokens[$i + 2][1] : $tokens[$i + 1][1];
                    $this->list[] = array("pos" => $i,"type" => $this->type[$tokens[$i][0]],"file" => trim($f, "\"\'"));
                }
            }
        }
    }

    public function find($find) {
        $finds = array_filter($this->list, function ($v) use($find) {
            return $v['file'] == $find;
        });

        return empty($finds) ? false : $finds;
    }

    public function getList() {
        return $this->list;
    }

    public function getFiles($type = null) {
        $finds = array_filter($this->list, function ($v) use($type) {
            return is_null($type) ? true : $type == $v['type'];
        });
        return empty($finds) ? false : $finds;
    }
}

Answer 2

Quoted from: How to know if php script is called via require_once()?

I was looking for a way to determine if a file have been included or called directly, all from within the file. At some point in my quest I passed through this thread. Checking various other threads on this and other sites and pages from the PHP manual I got enlightened and came up with this piece of code:

if (basename(__FILE__) == basename($_SERVER["SCRIPT_FILENAME"])) {
  echo "called directly";
} else {
  echo "included/required";
}

In essence it compares if the name of the current file (the one that could be included) is the same as the file that is beeing executed.

Credit: @Interwebs Cowboy

Answer 3

if (defined('FLAG_FROM_A_PARENT'))
// Works in all scenarios but I personally dislike this

if (__FILE__ == get_included_files()[0])
// Doesn't work with PHP prepend unless calling [1] instead.

if (__FILE__ == $_SERVER['SCRIPT_FILENAME'])
// May break on Windows due to mixed DIRECTORY_SEPARATOR

if (basename(__FILE__) == basename($_SERVER['SCRIPT_FILENAME']))
// Doesn't work with files with the same basename but different paths

if (realpath(__FILE__) == realpath($_SERVER['SCRIPT_FILENAME']))
// Seems to do the trick as long as document root is properly configured

Note: On WAMP Servers virtual-hosts sometimes inherit the default document root setting, causing $_SERVER['DOCUMENT_ROOT'] to display wrong path.

Answer 4

It's sooo simple.. I have made something like this:

//code for file.php
if (!isset($file_included)){
   echo "It was loaded!";
} else {
  echo "It was included!";
}

//code for loader.php
//proves that atleast loader.php has loaded,
//not the file we targeted first..
$file_included = true;
include("../file.php");

And that's it.. as simple as in python.

Answer 5

I took a similar approach to this issue when I cam across it. The solution I found was to load each file as needed in an include_once method. Hope this helps.

$FILES = get_included_files();  // Retrieves files included as array($FILE)
$FILE = __FILE__;               // Set value of current file with absolute path
if(!in_array($FILE, $FILES)){   // Checks if file $FILE is in $FILES
  include_once "PATH_TO_FILE";  // Includes file with include_once if $FILE is not found.
}

I have the following function established to check files loaded:

ARRAY_DUMP($FILES);

function ARRAY_DUMP($array){
  echo "
    <span style='font-size:12px;'>".date('h:i:s').":</span>
    <pre style='font-size:12px;'>", print_r($array, 1), "</pre>
  ";
}

Output:

currentArray
(
  [0] => /home/MY_DOMAIN/hardeen/index.php
  [1] => /home/MY_DOMAIN/hardeen/core/construct.php
  [2] => /home/MY_DOMAIN/hardeen/core/template.php
  [3] => /home/MY_DOMAIN/hardeen/bin/tags.php
  [4] => /home/MY_DOMAIN/hardeen/bin/systemFunction.php
)

Answer 6

<?php
    if (__FILE__ == $_SERVER['SCRIPT_FILENAME'])
    {
        //file was navigated to directly
    }
?>

Taken from mgutt's answer to a slightly different question here. It's important to note this doesn't work if the script is run from command line but other than that it functions exactly like python's

if __name__ == '__main__':

as far as I can tell