PHP URL包含不起作用

Question

I have a problem with an URL-include, which I don't understand...: For testing I have coded the following script:

<?php
error_reporting(E_ALL|E_STRICT);
ini_set('display_errors', 1);
echo "First text";
include("http://www.xxxxxxxxxx.de/includetest.php");
echo "Second text"; 
?>

Allow_url_include is set to on. (via php.ini)

Allor_url_fopen ist set to on. (via php.ini)

The includetest.php only contains plain text for testing. There is no php-code.

The result of that script is only the "first text". After that the script is stopped.

If I use "or die('not working');" after the include, the result is the whole text (also the second text) with the following warning:

Warning: include(1): failed to open stream: No such file or directory in /srv2/www/htdocs/xhtml-test/_baustelle/testphp02.php on line 6 Warning: include(): Failed opening '1' for inclusion (include_path='.:/usr/share/php:/usr/share/pear') in /srv2/www/htdocs/xhtml-test/_baustelle/testphp02.php on line 6

Why is that? I am at a loss...

Answer 1

Here is the problem of code:

// Won't work; file.txt wasn't handled by www.example.com as PHP
include 'http://www.example.com/file.txt?foo=1&bar=2';

// Won't work; looks for a file named 'file.php?foo=1&bar=2' on the
// local filesystem.
include 'file.php?foo=1&bar=2';

// Works.
include 'http://www.example.com/file.php?foo=1&bar=2';

Ref. of this Is Here

The file you are including is not a valid php file as it is already surved by a server as php.

This code should work as you want:

<?php
error_reporting(E_ALL|E_STRICT);
ini_set('display_errors', 1);
echo "First text";
echo file_get_contents("http://www.xxxxxxxxxx.de/includetest.php");
echo "Second text"; 
?>