www.youtubeinmp3.com/fetch/?format=JSON&video=http://www.youtube.com/watch?v=i62Zjga8JOM
示例输出:
{
"title":"Happy Forever Alone Day (Forever Alone Song)",
"length":"125",
"link":"http:'/'/youtubeinmp3.com'/download'/get' /?i=k1MakL%2FYlh6KUtEHqyQlg9XwtUUWO"
}
我想使用php-json文件获取内容,但这不起作用。
<?php
$json_url = "www.youtubeinmp3.com/fetch/?format=JSON&video=http://www.youtube.com/watch?v=i62Zjga8JOM";
$json = file_get_contents($json_url);
$data = json_decode($json, TRUE);
echo "<pre>";
print_r($data);
echo "</pre>";
?>
但没用,请帮帮我。。$json_ul="www.youtubeinmp3.com/fetch/?format=json&video=$videoid";$videoid是使用post变量获得的dyanamic值。
likie作为
<?php
$video=$_POST['yturl'];
echo $video;?>
谢谢。
使用此代码获取输出。
$curlSession = curl_init();
curl_setopt($curlSession, CURLOPT_URL, 'www.youtubeinmp3.com/fetch/?format=JSON&video=http://www.youtube.com/watch?v=i62Zjga8JOM&chof=json');
curl_setopt($curlSession, CURLOPT_BINARYTRANSFER, true);
curl_setopt($curlSession, CURLOPT_RETURNTRANSFER, true);
$jsonData = json_decode(curl_exec($curlSession));
curl_close($curlSession);
echo "<pre>";
print_r($jsonData);
echo "</pre>";