无法使用CURL获取Feed内容

Question

I'm trying to get the content of this feed : http://www.institut-viavoice.com/viavoice-paris/publications/sondages-publies?format=feed&type=rss

Here is my code :

$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;

I tried many other CURL options but it doesn't work. As the content is accessible through my browser, I suppose it can be done with PHP. But what is wrong with my code ? It seems like there is an exception with the server of this feed ?

Answer 1

Not sure, may be your breaking the cURL options and calling the URL. Here a simple example, give it a try:

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;