CODEWITHSUNDEEP
phpparsingurl
Yahoo
Yahoo

Reputation: 4187

Parsing a URL - Php Question

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

http://www.ewwsdf.org/012Q/rhod-05.php

How can i modify the code so that i get the output as

http://www.ewwsdf.org/012Q/

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 )

Upvotes: 0

Views: 1285

Answers (4)

CrayonViolent
CrayonViolent

Reputation: 32537

use pathinfo() instead, it shows relevant info already parsed

Upvotes: 3

Phil
Phil

Reputation: 165069

Just need the Directory name without the file name

Then use dirname(), eg

$lnk= "http://".$parse['host'].dirname($parse['path']);

Upvotes: 4

Kai Qing
Kai Qing

Reputation: 18843

you could do something like this:

$sections = explode("/", $_SERVER['REQUEST_URI']);
$folder = $sections[1];
$url = "http://www.ewwsdf.org/".$folder."/";

Upvotes: 1

NikiC
NikiC

Reputation: 101956

$lnk = "http://".$parse['host'].dirname($parse['path']).'/';

dirname returns the parent directory's path.

Upvotes: 3

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