How to Get Specified data from Curl?

Question

sorry if my english not very good,

i have a file, please check http://www.tvlangsung.com/jwplayer-drive/google.php

that google.php is

    <?php
$link = 'https://drive.google.com/file/d/0B1xQLLJtrzJoaWUxUHdqY01mRGM/view';
$api = 'https://api.blogit.vn/getlink.php?link='.$link.'&json=jwplayer';
$sources = curl($api);
function curl($url)
{
    $ch = @curl_init();
    curl_setopt($ch, CURLOPT_URL, $url);
    $head[] = "Connection: keep-alive";
    $head[] = "Keep-Alive: 300";
    $head[] = "Accept-Charset: ISO-8859-1,utf-8;q=0.7,*;q=0.7";
    $head[] = "Accept-Language: en-us,en;q=0.5";
    curl_setopt($ch, CURLOPT_USERAGENT, 'Mozilla/5.0 (Windows NT 6.1; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/37.0.2062.124 Safari/537.36');
    curl_setopt($ch, CURLOPT_ENCODING, 'gzip');
    curl_setopt($ch, CURLOPT_HTTPHEADER, $head);
    curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
    curl_setopt($ch, CURLOPT_SSL_VERIFYHOST, FALSE);
    curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, FALSE);
    curl_setopt($ch, CURLOPT_TIMEOUT, 60);
    curl_setopt($ch, CURLOPT_CONNECTTIMEOUT, 60);
    curl_setopt($ch, CURLOPT_FOLLOWLOCATION, TRUE);
    $page = curl_exec($ch);
    curl_close($ch);
    return $page;
}
?>

<?php echo $sources; ?>

the question is how can separated that data

so i can make a download button like

<a href="<?php echo $480p; ?>" target="_blank">Download 480p</a>
<a href="<?php echo $360p; ?>" target="_blank">Download 360p</a>

how can i take that redirector link for each quality ?

please help, im a newbie

Ray Hunter · Accepted Answer · 2017-02-19 18:51:28Z

2

You need to parse the results as json with the json_decode() method with your data. Then you can loop over the results. Remember the results are objects so you have to access them as such and not as arrays.

$data = json_decode($sources);

foreach($data as $file) {
  echo '<a href="' . $file->file . '" target="_blank">Download '. $file->label . 'p</a><br/>';
}

That should provide the links that you are looking to create.

answered Feb 19, 2017 at 18:51

Ray Hunter

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Castro Alhdo · Accepted Answer · 2017-02-19 18:45:10Z

1

You should parse the json file and generate button with the type attribute. It 's and array you should.

$array = json_decode($sources);
foreach($array as $s){
  if($a['type']=="mp4"){ 
    put your link code here
  }
}

answered Feb 19, 2017 at 18:45

Castro Alhdo

2913 silver badges8 bronze badges

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