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View Full Version : Getting the current pages url



p4plus2
03-12-2008, 12:34 AM
This is fairly simple based on what I hear, However I need to know if its possible to do that and get only the page your on without the file extension...

ex)
http://demo.com/randomdir/part_i_need.notthis

if this is possible to just get that part can somebody show me an example?


THANKS,
~p4plus2~

_Aerospace_Eng_
03-12-2008, 01:03 AM
Use http://us2.php.net/basename

p4plus2
03-12-2008, 10:16 PM
Thanks for the help!

for anybody who needs a similar code here it is:



<?php
$path = $_SERVER['REQUEST_URI'];
$extension = '.php';
$file = basename($path, $extension);
echo("$file");
?>


Or a more practical use(which is similar to how I am using it):



<?php
$path = $_SERVER['REQUEST_URI'];
$extension = substr(strrchr($path, '.'), 1);
$extension = "."."$extension";
$file = basename($path, $extension);
if($file == "index"){
echo('hello, and welcome to my site');
}
elseif($file == "random"){
echo('welcome to a random page!');
}
else{
die("unknown page");
}
?>

you can also change:


$extension = substr(strrchr($path, '.'), 1);
$extension = "."."$extension";


to this:


$extension = '.';
$extension .= substr(strrchr($path, '.'), 1);

just personal preference.



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