这很有助于获得标题
$imageurl = "http://p.twimg.com/AW2-vIUCAAE_1FN.png"; $ch = curl_init(); curl_setopt ($ch,CURLOPT_RETURNTRANSFER,1); curl_setopt ($ch,CURLOPT_URL,$image_url); curl_setopt ($ch,CURLOPT_CONNECTTIMEOUT,20); curl_setopt ($ch,CURLOPT_USERAGENT,$_SERVER['HTTP_USER_AGENT']); curl_setopt ($ch,CURLOPT_FOLLOWLOCATION,true); curl_setopt($ch,CURLOPT_HEADER,true); curl_setopt($ch,CURLOPT_NOBODY,true); $content = curl_exec ($ch); curl_close ($ch); print_r($content);
回国
HTTP/1.1 200 OK Server: Nginx X-peep-host: den2twapi014 Last-Modified: Mon,15 Aug 2011 03:41:50 GMT ETag: "7283847-dbb8-4aa830b1dabf0" Accept-Ranges: bytes Content-Length: 56248 Cache-Control: public,max-age=21600 Expires: Wed,28 Mar 2012 15:53:57 GMT Content-Type: image/png X-CDN: AKAM Date: Wed,28 Mar 2012 09:53:57 GMT Connection: keep-alive
回应Content-Type的最佳方式是什么？
试试这个：-
<?PHP $image_url = "http://p.twimg.com/AW2-vIUCAAE_1FN.png"; $ch = curl_init(); curl_setopt ($ch,true); $content = curl_exec ($ch); $contentType = curl_getinfo($ch,CURLINFO_CONTENT_TYPE); echo $contentType; echo "<br/>"; curl_close ($ch); echo "<pre>"; print_r($content); ?>