Parsing a URL - Php Question
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Adi Mathur
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Published on 2011-01-31T23:21:38Z
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2011/01/31
23:26 UTC
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I am using the Following code
<?php
$url = 'http://www.ewwsdf.org/012Q/rhod-05.php?arg=value#anchor';
$parse = parse_url($url);
$lnk= "http://".$parse['host'].$parse['path'];
echo $lnk;
?>
This is giving me the output as
How can i modify the code so that i get the output as
Just need the Directory name without the file name
( I actually need the link so that i can link up the images which are on the pages, By appending the link behind the image Eg http://www.ewwsdf.org/012Q/hi.jpg )
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