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CURL用来从一些网页中获取内容。我需要从内容中提取媒体标签。

有没有可用的图书馆?或者任何关于制作它的想法都会非常棒。

4

1 回答 1

1

这会有帮助吗?

function file_get_contents_curl($url)
{
    $ch = curl_init();

    curl_setopt($ch, CURLOPT_HEADER, 0);
    curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
    curl_setopt($ch, CURLOPT_URL, $url);
    curl_setopt($ch, CURLOPT_FOLLOWLOCATION, 1);

    $data = curl_exec($ch);
    curl_close($ch);

    return $data;
}

$html = file_get_contents_curl("http://example.com/");

//parsing begins here:
$doc = new DOMDocument();
@$doc->loadHTML($html);
$nodes = $doc->getElementsByTagName('title');

//get and display what you need:
$title = $nodes->item(0)->nodeValue;

$metas = $doc->getElementsByTagName('meta');

for ($i = 0; $i < $metas->length; $i++)
{
    $meta = $metas->item($i);
    if($meta->getAttribute('name') == 'description')
        $description = $meta->getAttribute('content');
    if($meta->getAttribute('name') == 'keywords')
        $keywords = $meta->getAttribute('content');
}

echo "Title: $title". '<br/><br/>';
echo "Description: $description". '<br/><br/>';
echo "Keywords: $keywords";

或者,如果您需要保存图像..

$remote_img = 'http://www.example.com/images/image.jpg ';
$img = imagecreatefromjpeg($remote_img);
$path = 'images/';
imagejpeg($img, $path);

function save_image($img,$fullpath){
    $ch = curl_init ($img);
    curl_setopt($ch, CURLOPT_HEADER, 0);
    curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
    curl_setopt($ch, CURLOPT_BINARYTRANSFER,1);
    $rawdata=curl_exec($ch);
    curl_close ($ch);
    if(file_exists($fullpath)){
        unlink($fullpath);
    }
    $fp = fopen($fullpath,'x');
    fwrite($fp, $rawdata);
    fclose($fp); 
}
于 2012-08-13T00:25:35.733 回答