Easy way to get Vimeo id from a vimeo url

Question

I'm trying to get just the id from a vimeo URL. Is there a simpler way than this? All the vimeo video urls I see are always:

https://vimeo.com/29474908

https://vimeo.com/38648446

// VIMEO   $vimeo = $_POST['vimeo'];  function getVimeoInfo($vimeo) {     $url = parse_url($vimeo);     if($url['host'] !== 'vimeo.com' &&             $url['host'] !== 'www.vimeo.com')         return false;    if (preg_match('~^http://(?:www\.)?vimeo\.com/(?:clip:)?(\d+)~', $vimeo, $match))     {        $id = $match[1];    }    else    {        $id = substr($link,10,strlen($link));    }     if (!function_exists('curl_init')) die('CURL is not installed!');    $ch = curl_init();    curl_setopt($ch, CURLOPT_URL, "http://vimeo.com/api/v2/video/$id.php");    curl_setopt($ch, CURLOPT_HEADER, 0);    curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);    curl_setopt($ch, CURLOPT_TIMEOUT, 10);    $output = unserialize(curl_exec($ch));    $output = $output[0];    curl_close($ch);    return $output['id']; }  $vimeo_id = getVimeoInfo($vimeo); 

Answer 1

There are lot many vimeo URLs that are valid. Few examples are

All valid URLs:

http://vimeo.com/6701902 http://vimeo.com/670190233 http://player.vimeo.com/video/67019023 http://player.vimeo.com/video/6701902 http://player.vimeo.com/video/67019022?title=0&byline=0&portrait=0 http://player.vimeo.com/video/6719022?title=0&byline=0&portrait=0 http://vimeo.com/channels/vimeogirls/6701902 http://vimeo.com/channels/vimeogirls/67019023 http://vimeo.com/channels/staffpicks/67019026 http://vimeo.com/15414122 http://vimeo.com/channels/vimeogirls/66882931 

All invalid URLs:

http://vimeo.com/videoschool http://vimeo.com/videoschool/archive/behind_the_scenes http://vimeo.com/forums/screening_room http://vimeo.com/forums/screening_room/topic:42708 

I wrote this java regex that catches all the above valid URLs and rejects the invalid ones. I m not sure though if they vimeo has more valid URLs.

(https?://)?(www.)?(player.)?vimeo.com/([a-z]*/)*([0-9]{6,11})[?]?.* 

Hope this helps...

Answer 2

I think using parse_url() is the best option:

$vimeo = 'https://vimeo.com/29474908';  echo (int) substr(parse_url($vimeo, PHP_URL_PATH), 1);