How do I load a PHP file into a variable?

Question

I need to load a PHP file into a variable. Like include();

I have loaded a simple HTML file like this:

$Vdata = file_get_contents(\"text         


        

Answer 1

Alternatively, you can start output buffering, do an include/require, and then stop buffering. With ob_get_contents(), you can just get the stuff that was outputted by that other PHP file into a variable.

Answer 2

I suppose you want to get the content generated by PHP, if so use:

$Vdata = file_get_contents('http://YOUR_HOST/YOUR/FILE.php');

Otherwise if you want to get the source code of the PHP file, it's the same as a .txt file:

$Vdata = file_get_contents('path/to/YOUR/FILE.php');

Answer 3

If you want to load the file without running it through the webserver, the following should work.

$string = eval(file_get_contents("file.php"));

This will load then evaluate the file contents. The PHP file will need to be fully formed with <?php and ?> tags for eval to evaluate it.

Answer 4

file_get_contents() will not work if your server has allow_url_fopen turned off. Most shared web hosts have it turned off by default due to security risks. Also, in PHP6, the allow_url_fopen option will no longer exist and all functions will act as if it is permenantly set to off. So this is a very bad method to use.

Your best option to use if you are accessing the file through http is cURL

Answer 5

ob_start();
include "yourfile.php";
$myvar = ob_get_clean();

ob_get_clean()

Answer 6

If your file has a return statement like this:

<?php return array(
  'AF' => 'Afeganistão',
  'ZA' => 'África do Sul',
  ...
  'ZW' => 'Zimbabué'
);

You can get this to a variable like this:

$data = include $filePath;