Capturing Part Of URL

[Hate]

Hello,

 

I'm trying to capture the red part of the url. The blue part of the URL I would like to be optional. Case should not matter. Should I have the "http://" part a second optional as well? Or should I put it with "www."? Ideally, I just need to ensure that it pulls the video title from that domain regardless of how they accessed the website.  What would be the best possible way to achieve this with regex?

 

Summarized: I just need whatever is after /videos/ but would ideally like to ensure that the domain is correct as well.

 

http://www.cooltools.org/videos/how-to-style-menu/

 

Thanks.

[MarPlo]

Hi,

I think this code is good for what you want:

$url = 'http://www.cooltools.org/videos/how-to-style-menu/';
if(preg_match('#(http://){0,1}(www.){0,1}cooltools.org/videos/([^/]+)#i', $url, $mc)) {
  $video = $mc[3];
  echo $video;         // how-to-style-menu
}

[xyph]

You don't really need RegEx to do this. Here's a couple methods, one involving a VERY simple RegEx.

 

<?php

$url = 'http://www.cooltools.org/videos/how-to-style-menu/';

$parsed = parse_url($url);

print_r($parsed);
/*
returns:
Array
(
    [scheme] => http
    [host] => www.cooltools.org
    [path] => /videos/how-to-style-menu/
)
*/

$paths = explode( '/', $parsed['path'] );

foreach( $paths as $path ) {
if( !empty($path) ) { // ignore empty results from leading/trailing slashes
	echo '<h3>'.$path.'</h3>';
}
}

// Alternate method to explode, uses RegEx. Allows empty splits to be ignored automatically
$paths = preg_split( '#/#', $parsed['path'], NULL, PREG_SPLIT_NO_EMPTY );
foreach( $paths as $path ) {
echo '<h3>'.$path.'</h3>';
}

?>

[requinix]

And there's yet another method. Since URLs look kinda like file paths,

basename(trim($url, "/"))

(Of course this doesn't validate the rest of the URL. Only grabs the last component of the path.)