Parsing a URL - Php Question

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 )

In this question has a brief sample using Zend < stackoverflow.com/questions/4719084/… > — Felipe Cardoso Martins
– Felipe Cardoso Martins, Commented Jan 31, 2011 at 23:46

Phil · Accepted Answer · 2011-01-31 23:25:15Z

4

Just need the Directory name without the file name

Then use dirname(), eg

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

answered Jan 31, 2011 at 23:25

Phil

166k25 gold badges265 silver badges269 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Phil Over a year ago

@Adi To be fair, @nikic gave this same answer first (by 13 seconds)

NikiC · Accepted Answer · 2011-01-31 23:25:02Z

3

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

dirname returns the parent directory's path.

answered Jan 31, 2011 at 23:25

NikiC

102k39 gold badges194 silver badges226 bronze badges

Comments

CrayonViolent · Accepted Answer · 2011-01-31 23:26:26Z

3

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

answered Jan 31, 2011 at 23:26

CrayonViolent

32.6k6 gold badges61 silver badges80 bronze badges

3 Comments

mfonda Over a year ago

+1. $parse = pathinfo($url); echo $parse['dirname']; is exactly what you want.

NikiC Over a year ago

@mfonda: No, it isn't he would need to first parse_url and then pathinfo the 'path'. It is better in this case to use just dirname.

CrayonViolent Over a year ago

@nikic: why would he need to first use parse_url()? pathinfo() can take a full URL as an argument

Kai Qing · Accepted Answer · 2011-01-31 23:25:06Z

1

you could do something like this:

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

answered Jan 31, 2011 at 23:25

Kai Qing

18.9k6 gold badges41 silver badges61 bronze badges

Collectives™ on Stack Overflow

Parsing a URL - Php Question

4 Answers 4

1 Comment

Comments

3 Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

1 Comment

Comments

3 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related