how to i get only URL in json in PHP -


how data in json of mention format want data formate of json
{"22":{"quality":"22","type":"video\/mp4","url":"http://
please guide me how can this.
in advance

i have try code

$url = 'http://api.miyulasi.com/youtube/1.0.0/download.php?id=7lcdeyxw3mm'; $data = file_get_contents($url); header('content-type: application/json'); echo $data;

and responde this

<div class="col-md-3 col-sm-3 col-xs-6 text-center downbuttonbox"><a href="http://redirector.googlevideo.com/videoplayback?

here go

<?php  $url = 'http://api.miyulasi.com/youtube/1.0.0/download.php?id=7lcdeyxw3mm'; $data = file_get_contents($url);  $doc = new domdocument(); @$doc->loadhtml($data); $xpath = new domxpath($doc);  $links = array(); $expression = "//a[contains(@class, 'btn btn-default btn-sm downbuttonstyle')]"; foreach($xpath->evaluate($expression) $link) {     $content = $link->textcontent;     $link = $link->getattribute('href');     preg_match("/&mime=([^&]*)&/", $link, $mime);     preg_match("/&itag=([^&]*)&/", $link, $itag);     preg_match("/\((.*)\)/", $content, $quality);      $links[] = array("itag" => $itag[1], "type" => $mime[1], "url" => $link, "quality" => $quality[1]); } header('content-type: application/json'); echo json_encode($links, json_pretty_print);

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