+1 vote
by
There is a script that outputs the content depending on the url. What code do I need to display the content of filter33.php?
<?php
$rurl = $_SERVER["REQUEST_URI"];
$pagename="";
if ($rurl=="/filter/" ) {$pagename="Тут вывести содержание файла /filter/33.php ";}

echo"$pagename";
?>
Спасибо
by
file_get_contents() ?
I'm a little confused by the question.
by
Vladislav , show me a non-nonsense solution to this problem
by
This script outputs text, html code only on the right Url without any problems, but I need to output code from an external file on the right url.
by
Who teaches you such nonsense.....

1 Answer

0 votes
by
 
Best answer
<?php
if ($_SERVER["REQUEST_URI"] == "/filter/") {
require_once $_SERVER["DOCUMENT_ROOT"]."/filter/33.php";
}
?>
...