Jump to content

php error: Failed to Open Stream


samg

Recommended Posts

You are including a file from your own server, right?

 

Otherwise, the server you're trying to send the request to might have set header('Access-Control-Access-Origin: http://owndomain.com'); so that only their own domain can send HTTP requests to their site and get responses.

Just an educated guess, though.

Thanks for the quick response.

 

@Svenskunganka :     I tried to post my code but it doesn't allow me :(. As soon as I click save, it disappears :( . Let me try again.

@Irate: Yes I am trying to include a file on my sever itself.

 

1. Here is the dir structure:

a) /test/foo.php --> this has include to my own server. I tried to paste the exact code but it removed it :(. bascially its pointing to my own server using SERVER variables pointing to "bar.php"

b) /test/bar.php

 

2. Apache document root is pointing to "/test" like

=====

/var/www/html --> /test

=====

 

3. echo __DIR__  shows me "/test" so its definitely pointing to the right directory.

4. I have given full permssiion to this directory, in case that's the issue but no luck.

5. Exact error in apache error log. I had to remove the exact URL as this post doesn't allow me post that.

=====

[Mon Jul 08 10:41:49 2013] [error] [client ] PHP Warning:  include<URL>: failed to open stream: HTTP request failed!  in /test/foo.php

[Mon Jul 08 10:39:49 2013] [error] [client ] PHP Warning:  include(): Failed opening <URL> for inclusion (include_path='.:/usr/share/pear:/usr/share/php') in /test/foo.php

=====

 

Let me know if any other info is needed.

 

Thanks

I assume you mean multiple includes in the same file.  If so, then two ways.  One is what you just described, but if you use a function you only need one include:

//foo.php
$a=1;
$b=2;
include('bar.php');

$a=3;
$b=4;
include('bar.php');

//bar.php
echo $a+$b;

Or using a function:

//foo.php
include('bar.php');
show_addition(1, 2);
show_addition(3, 4);

//bar.php
function show_addition($arg1, $arg2) {
   echo $arg1+$arg2;
}

Archived

This topic is now archived and is closed to further replies.

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.