最近要用到通过post上传文件,网上盛传的有curl的post提交和fsockopen,其中curl最简单,于是从最简单的说起。
这是简单的将一个变量post到另外一个页面
$url = ''; $data = array('a'=> 'b'); $ch = curl_init(); curl_setopt($ch, CURLOPT_URL, $url); curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1); curl_setopt($ch, CURLOPT_POST, true); curl_setopt($ch, CURLOPT_POSTFIELDS, $data); $ret = curl_exec($ch); curl_close($ch);
主要说下这个选项CURLOPT_RETURNTRANSFER:如果设置为true/1,则curl_exec的时候不会自动将请求网页的内容输出到屏幕,$ret为请求网页的内容,如果设置为false/0,则curl_exec的时候会自动将请求网页的内容输出到屏幕,此时如果请求成功的话$ret的内容是1或者true。
下面是上传本地文件的代码,如果需要上传远程文件,则先down到本地,然后删掉即可(如有同学有别的办法还请告知):
$url = ''; $file = '1.jpg'; $field['uploadFile'] = '@'.$file;(uploadFile为接收端的name名) $ch = curl_init(); curl_setopt($ch, CURLOPT_URL, $url); curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1); curl_setopt($ch, CURLOPT_POST, 1); curl_setopt($ch, CURLOPT_POSTFIELDS, $field); $ret = curl_exec($ch); curl_close($ch);
这是fsockopen的办法: $uploadInfo = array( 'host'=>'', 'port'=>'80', 'url'=>'/upload.php' ); $fp = fsockopen($uploadInfo['host'],$uploadInfo['port'],$errno,$errstr); $file = '1.jpg'; $content = file_get_contents($file); $boundary = md5(time()); $out.="--".$boundary." "; $out.="Content-Disposition: form-data; name="uploadFile"; filename="".$file."" "; $out.="Content-Type: image/jpg "; $out.=$content." "; $out.="--".$boundary." "; fwrite($fp,"POST ".$uploadInfo['url']." HTTP/1.1 "); fwrite($fp,"Host:".$uploadInfo['host']." "); fwrite($fp,"Content-Type: multipart/form-data; boundary=".$boundary." "); fwrite($fp,"Content-length:".strlen($out)." "); fwrite($fp,$out); while (!feof($fp)){ $ret .= fgets($fp, 1024); } fclose($fp); $ret = trim(strstr($ret, " ")); preg_match('/http:.*/', $ret, $match); return $match[0];