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code:inside Feed.php

$language = 'en';
include_once '../lang/Feed_' .$language. '.php';

 

error:

Warning: showfeed(../lang/Feed_en.php) [function.showfeed]: failed to open stream: No such file or directory in /home/frdowd3/public_html/alex/alpha/forms/Feed.php on line 5

Warning: showfeed() [function.include]: Failed opening '../lang/Feed_en.php' for inclusion (include_path='.:/usr/lib/php:/usr/local/lib/php') in /home/frdowd3/public_html/alex/alpha/forms/Feed.php on line 5

 

directory structure:

alpha\
-forms\
          -Feed.php
-lang\
         -Feed_en.php

 

what can I do to fix the error, and I have checked all files are where they are suppose to be.

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https://forums.phpfreaks.com/topic/76886-solved-gives-problems/
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A couple of things: When a file (say dir1/foo.php) is included into another file (say dir2/foo.php), the source must be relative too the foo.php in dir2. Second perhaps the server does not have permission to execute the file? And possibly third does the file exist?

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https://forums.phpfreaks.com/topic/76886-solved-gives-problems/#findComment-389554
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Very Simple because include(); is a function it needs to be inside brackets:

Change

<?php
$language = 'en';
include_once '../lang/Feed_' .$language. '.php';
?>

To...

<?php
$language = 'en';
include_once ('../lang/Feed_' .$language. '.php');
?>

 

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https://forums.phpfreaks.com/topic/76886-solved-gives-problems/#findComment-389585
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