How to get mp4 url on source page? - php

Question

My page source have this html (js):
<script>
playlist.sources.push({
label: "480p",
source: "//zingtv-video-14.zadn.vn/Video480/2016/0418/7c/1f1fd73f7432d7524cb43b57da35d6df.mp4?authen=exp=1559094603~acl=1f1fd73f7432d7524cb43b57da35d6df~hmac=01da84812b2ef7d43e55be79fa3ef56e",
index: 1
});</script>
I want to get:
//zingtv-video-14.zadn.vn/Video480/2016/0418/7c/1f1fd73f7432d7524cb43b57da35d6df.mp4?authen=exp=1559094603~acl=1f1fd73f7432d7524cb43b57da35d6df~hmac=01da84812b2ef7d43e55be79fa3ef56e
My code :
<?php
// Check if the URL parameter for our proxy is set.
if (!empty($_GET['url'])) {
if (filter_var($_GET['url'], FILTER_VALIDATE_URL)) {
$grabs = file_get_contents("{$_GET['url']}");
$grab = json_decode($grabs, true);
// I don't know what to do next
echo "{$video}" />";
} else {
echo "Given URL is not valid.";
}
} else {
echo "You need to specify the URL.";
}
?>
and show in the content page.
How do I solve this problem?

Answer 1

Here, we can try using preg_match_all with a simple expression:
source: "(.+)",
where our desired output is saved in a capturing group $1.
Test
$re = '/source: "(.+)",/m';
$str = '<script>
playlist.sources.push({
label: "480p",
source: "//zingtv-video-14.zadn.vn/Video480/2016/0418/7c/1f1fd73f7432d7524cb43b57da35d6df.mp4?authen=exp=1559094603~acl=1f1fd73f7432d7524cb43b57da35d6df~hmac=01da84812b2ef7d43e55be79fa3ef56e",
index: 1
});</script>';
preg_match_all($re, $str, $matches, PREG_SET_ORDER, 0);
var_dump($matches[0][1]);
DEMO
I'm not quite sure if this file_get_content would work:
$str = file_get_contents('https://tv.zing.vn/video/id/IWZBFE8U.html?t=388');
$str = mb_convert_encoding($str, 'HTML-ENTITIES', "UTF-8");
var_dump($str);