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);
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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.="--".$boundary."\r\n";$out.="Content-Disposition: form-data; name=\"uploadFile\"; filename=\"".$file."\"\r\n";$out.="Content-Type: image/jpg\r\n\r\n";$out.=$content."\r\n";$out.="--".$boundary."\r\n"; fwrite($fp,"POST ".$uploadInfo[‘url‘]." HTTP/1.1\r\n");fwrite($fp,"Host:".$uploadInfo[‘host‘]."\r\n");fwrite($fp,"Content-Type: multipart/form-data; boundary=".$boundary."\r\n");fwrite($fp,"Content-length:".strlen($out)."\r\n\r\n");fwrite($fp,$out);while (!feof($fp)){ $ret .= fgets($fp, 1024);}fclose($fp);$ret = trim(strstr($ret, "\r\n\r\n"));preg_match(‘/http:.*/‘, $ret, $match);return $match[0];
时间: 2024-10-10 23:13:23