include(""); --> Warning: Failed opening '' for inclusion

[mlara]

Hi

I have the follow problem:

server 1: mypage.php
server 2: pageserver2.php

in mypage.php I write:

<?php include(&quot;

http://x.y.z/pageserver2.php?var1=help&quot;);

and then I get this message:

Warning: Failed opening '

http://x.y.z/pageserver2.php?var1=help'

for inclusion (include_path='.:/php/includes:/usr/share/php') in /htdocs/sam/mypage.php on line #

Somebody could help me?

Thanks.

 

[sleipnir214]

Normally, include() requires a filename on the local filesystem, not a web address.

If and only if you are running PHP version 4.3.0 or higher and you have the runtime configuration directive &quot;allow_url_fopen&quot; can you use a URL with include().

Please see

http://www.php.net/manual/en/function.include.php

for more information, particularly paying attention to the Warning box on that page. Want the best answers? Ask the best questions:

http://www.tuxedo.org/~esr/faqs/smart-questions.html

TANSTAAFL!

 

[sleipnir214]

Correction. The second paragraph should have read '...and you have the runtime configuration directive &quot;allow_url_fopen&quot; set to &quot;on&quot; can you use...' Want the best answers? Ask the best questions:

http://www.tuxedo.org/~esr/faqs/smart-questions.html

TANSTAAFL!

 

[youradds]

You tried;

<?php include(&quot;

http://x.y.z/pageserver2.php?var1=help&quot;);

?>

(note there is no ; after the &quot now )

or...

<?php include(&quot;

http://x.y.z/pageserver2.php&quot;);

?>

to see if the ?var1=help is the problem.

Cheers


Andy