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Home/ Questions/Q 971179
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Editorial Team
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Editorial Team
Asked: May 16, 20262026-05-16T02:55:14+00:00 2026-05-16T02:55:14+00:00

When I link to a file in a directory, I want to open that

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When I link to a file in a directory, I want to open that file as a param in a php program.

For example, I have a directory temp, and files aa.txt, bb.txt, and test.php. When I link to aa.txt it should be handled like test.php?f=aa.txt.

What do I change in the .htaccess file?

the code in the test.php

<?php
$f=$_GET['f'];
if(@file_exists($f)){
    $inhoud = file_get_contents($f);
}else{
    $inhoud = "Not found\n";
}
print "hallo <hr>".$inhoud;

?>
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-16T02:55:15+00:00Added an answer on May 16, 2026 at 2:55 am

    You want to use mod_rewrite, and define rules like the following within a .htaccess file in the directory you want it to apply to:

    # Enable mod_rewrite
    RewriteEngine on
    
    # If the request is an actual file and not test.php...
    RewriteCond %{REQUEST_FILENAME} !test.php$
    RewriteCond %{REQUEST_FILENAME} -f
    
    # ... then rewrite the URL to pass the filename as a parameter to test.php
    RewriteRule /(.*)$ test.php?f=$1 [QSA]
    
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