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572597 Symfony 2 Adult Video Site PHP


I'm interested in purchasing a bit of code. Right now I have a Symfony 2-based site up and running and I'd like to add the information I need for every (or as many as possible) adult site.

All I really need is this information:

$regexes[] = '(?:https?:\/\/(?:.+\.)[url removed, login to view](?:\/video\/|(?:.+clip_id=))([a-z0-9-]+))';

and this information:

//vimeo video ids will go on matches[2]

foreach (array_unique($matches[2]) as $videoId) {

if (empty($videoId)) {



//get video data

$duration = 0;

$videoDataUrl = "[url removed, login to view]{$videoId}.php";

$videoData = unserialize(file_get_contents($videoDataUrl));

if (!empty($videoData[0])) {

$thumbUrl = $videoData[0]['thumbnail_large'];

$videoUrl = $videoData[0]['url'];

//type 2 now hardcoded, we should create a VideoResourceFetcher

$urls[] = array('type' => 2, 'url' => $thumbUrl, 'src_th_url' => $videoUrl, 'duration' => $duration);


...but for every adult video site in existence. (Or at least an even dozen of them.)

I don't need it to download videos for me... I don't need it to do anything at all really. I just need that information for all the adult sites.

Let me know if you think you can help.

Kĩ năng: Bất kì công việc gì, PHP, Dịch vụ video, Thiết kế trang web, Quản lý website, XXX

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( 0 nhận xét ) Spain

ID dự án: #2318568