file_get_content不适用于特殊网站

Question

我希望得到https://06fazmusic.com/的内容，但file_get_content不起作用。

$context = stream_context_create(array(
'http' => array(
    'method' => "GET",
    'follow_location' => false,
    'header' => "Accept-Language: en-US,en;q=0.8rn" .
        "Accept-Encoding: gzip,deflate,sdchrn" .
        "Accept-Charset:UTF-8,*;q=0.5rn" .
        "Accept-Language:en-US,en;q=0.8" .
        "Cache-Control:max-age=0" .
        "User-Agent:Mozilla/5.0 (Windows NT 10.0; Win64; x64) 
AppleWebKit/537.36 (KHTML, like Gecko) Chrome/63.0.3239.132 Safari/537.36"
    )
));


$post_url='https://06fazmusic.com/omid-called-namaze-eshgh/';
$array = get_headers($post_url);
echo file_get_contents($post_url, false, $context);  

Answer 1

我没有使用file_get_contents（）。我使用Curl类型来获取该页面的代码是：

function curl_get_contents($url)
{
  $ch = curl_init();
  curl_setopt($ch, CURLOPT_URL, $url);
  curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
  curl_setopt($ch, CURLOPT_FOLLOWLOCATION, 1);
  $data = curl_exec($ch);
  curl_close($ch);
  return $data;
}
echo curl_get_contents($post_url);

它正在工作输出屏幕截图是：