質問

If I were using youtube for an example, $link = "http://www.youtube.com/watch?v=hYVVCRqTz1Q"

In the source code of http://www.youtube.com/watch?v=hYVVCRqTz1Q we can find

    <meta property="og:image" content="http://i1.ytimg.com/vi/hYVVCRqTz1Q/hqdefault.jpg?feature=og">

I would like php to use $link to obtain $thumbnail where

$thumbnail = "http://i1.ytimg.com/vi/hYVVCRqTz1Q/hqdefault.jpg"
役に立ちましたか?

解決

$link = 'http://www.youtube.com/watch?v=hYVVCRqTz1Q';

Now match if the URL is a valid Youtube Link by matching it with Pattern.

$ypattern= "#(http://www.youtube.com)?/(v/([-|~_0-9A-Za-z]+)|watch\?v\=([-|~_0-9A-Za-z]+)&?.*?)#";
if(preg_match_all($ypattern,$link, $output))
{ 
    foreach ($output[4] AS $video_id)
    {

Youtube Headers

$headers = get_headers('http://gdata.youtube.com/feeds/api/videos/' . $video_id)`;

     if (strpos($headers[0], '200'))
     {
    $youtube = simplexml_load_file('http://gdata.youtube.com/feeds/api/videos/'.$video_id.'?v=1');
            $json = json_decode(file_get_contents('http://gdata.youtube.com/feeds/api/videos/'.$video_id.'?v=2&alt=jsonc'));

Thumbnail >

$thumbnail = $json->data->thumbnail->sqDefault;
                }
            }
        }
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