无法获得饲料内容，卷曲

Question

我试图让这个饲料的内容： http://www.institut-viavoice.com/viavoice-paris/publications/sondages-publies?format=feed&type=rss

这里是我的代码：

$url = 'http://www.institut-viavoice.com/viavoice-paris/publications/sondages-publies?format=feed&type=rss'; 

$options = array(
CURLOPT_RETURNTRANSFER => true, 
CURLOPT_HEADER   => false, 
CURLOPT_FOLLOWLOCATION => true, 
CURLOPT_ENCODING  => "", 
CURLOPT_USERAGENT  => "Mozilla/5.0 (Windows NT 6.1; WOW64; rv:18.0) Gecko/20100101 Firefox/18.0", 
CURLOPT_AUTOREFERER => true, 
CURLOPT_CONNECTTIMEOUT => 30, 
CURLOPT_TIMEOUT  => 30, 
CURLOPT_MAXREDIRS  => 10 
); 

$curl = curl_init($url); 
curl_setopt_array($curl, $options); 
$content = curl_exec($curl); 
curl_close($curl); 

echo $content; 

我试过很多其他卷曲选择，但它不”工作。由于内容可以通过我的浏览器访问，我想可以用PHP来完成。但是我的代码有什么问题？这个提要的服务器似乎有个例外？

Answer 1

不确定，可能是您打破cURL选项并调用URL。 这里有一个简单的例子，试试看：

function get_data($url) { 
    $ch = curl_init(); 
    $timeout = 5; 
    curl_setopt($ch, CURLOPT_URL, $url); 
    curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1); 
    curl_setopt($ch, CURLOPT_CONNECTTIMEOUT, $timeout); 
    curl_setopt($ch, CURLOPT_USERAGENT, "Mozilla/5.0 (Windows NT 6.1; WOW64; rv:18.0) Gecko/20100101 Firefox/18.0"); 
    $data = curl_exec($ch); 
    curl_close($ch); 
    return $data; 
} 

$content = get_data('http://www.institut-viavoice.com/viavoice-paris/publications/sondages-publies?format=feed&type=rss'); 

echo $content;