PHP variables maintaining an URL string.

Question

I'm not a PHP developer, but I'm currently hacking on an internal tool so my team can take advantage of its goodness. There's an index file that looks like so:

require( ($loader_path = "../../loaderapi/") . "loader.php" );

Used like this, $loader_path will retain its value within the loader.php file.

However, we want to access this API from our team's server like so:

require( ($loader_path = "http://remoteservername/loaderapi/") . "loader.php" );

In this case the $loader_path variable doesn't retain its value. I'm guessing it has something to do with it being a full blown URL, but I might be wrong. Any idea on how I can make this work, or why I can't do it this way?

Answer 1

Have you tried splitting it into two lines:

$loader_path = "http://remoteservername/loaderapi/";
require( $loader_path . "loader.php" );

It's easier to read this way as well.

Answer 2

If your accessing a PHP script over HTTP, only the output of that script is returned. So your script will try to interpret the output of that remote PHP script and not its source.

If there is a connection over the file system, you may want to try file://remoteservername/loaderapi/loader.php instead.

Answer 3

Simplify the code reading by simply putting everything on 3 lines:

$loader_path = "http://remoteservername/loaderapi/";
$page = "loader.php";
require($loader_path . $page );

Much clearer and it works.

Answer 4

why not just put it above the require statement? would make it easier to read too.

<?php
  $loader_path = "../../folderName/"
  require($loader_path . "filename")
 ?>

Answer 5

There shouldn't be any real difference between the two; what you're doing is defining $loader_path, concatenating the loader.php, and passing that to require.

HOWEVER: you're defining the variable within the scope of a require, which will halt processing of the script of require fails.

Try replacing 'require' with 'include' and see if it retains the variable.

Also, note that if you are running your PHP server on a windows machine, and the php version is less than 4.3.0, neither 'require' nor 'include' can handle remote files : http://us.php.net/manual/en/function.include.php

Also, as noted before, if the .php lives on a remote server that parses php, you will not get code, but the result of the remote server processing the code. You'll either have to serve it up as a .txt file, or write php that, when processed, outputs valid php.

Answer 6

NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO!

Remote file inclusion is a BAD idea, probably one of the biggest security flaws you can open up. Even for an internal tool this is not acceptable even if only purely for contributing bad habits.

PHP by default disables this behavior, and there is a broad push to have the ability to perform an include on a URL completely stripped from PHP (as there is no compelling reason to have this ability).

If you want to load shared resources, go through a shared file system drive (as in, don't use http, ftp, anything but file://) or better yet distribute copies of loader.php through a version control system. Loading from a single file resource opens you up to problems in the future of say a new dev overwriting loader.php and breaking everyone else's code.