samg

If I have multiple "include" for the same file with same parameters but with different values for those parameters, should I re-initialize my parameters before each include? Is that the way to go?

I may have figured out what the problem may be. I am passing query parameters to the file in include and I guess it considers those parameters as part of file name. Do you know how I can pass query parameters in the "include"?

@Shawn: with that I get, No such file or directory even though "echo __DIR__" returns "test" which contains "bar.php"

@AbraCadaver: Yep. that property is already set to "On".

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

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Hi, I have the "include" in my code but it errors out with "failed to open stream: HTTP request failed! " error. I already have this set in php.ini "allow_url_include = On" but it still fails. Can anybody help? Thanks