PHP页面跳转与跨站提交伪造Referer地址来源
一、尝试过的URL跳转方法,代码如下:
- echo '<meta http-equiv="refresh" content="0; URL='.$url.'">';
-
- echo '<scrīpt language="Javascrīpt">window.location.href="'.$url.'";</scrīpt>';
-
- echo '<script language="Javascrīpt">window.location.replace="'.$url.'";</ script>';
以上三种方法均无法传递REFERER地址.
二、使用PHP Socket函数伪造REFER
下面是PHP伪造REFERER代码部分,经过测试可以实现REFERER地址传递,其中$url是输入地址,代码如下:
- $uinfo = parse_url($url);
-
- if($uinfo['path'])
-
- $data = $uinfo['path'];
-
- else
-
- $data = '/';
-
- if(!$fsp = @fsockopen($uinfo['host'], (($uinfo['port']) ? $uinfo['port'] : "80"), $errno, $errstr, 12)){
-
- echo "对不起对方网站暂时无法打开,请您稍后访问:".$uinfo['host']; exit;
-
- }else{
-
- fputs($fsp, "GET “.$data .” HTTP/1.0rn");
-
- fputs($fsp, "Host: ".$uinfo['host']."rn");
-
- fputs($fsp, "Referer: phpfensi.comrn");
-
- fputs($fsp, "User-Agent: Mozilla/4.0 (compatible; MSIE 6.0; Windows NT 5.1)rnrn");
-
- $res='';
-
- while(!feof($fsp)) {
-
- $res.=fgets($fsp, 128);
-
- if(strstr($res,"200 OK")) {
-
- header("Location:$url"); exit;
-
- }
-
- }
-
- }
-
-
-
-
-
- $arr=explode("n",$res);
-
- $arr=explode(": ",$arr[3]);
-
- header("location:".$arr[0]);
-
- exit;
利用另一种方法 curl)伪造HTTP_REFERER,代码如下:
-
- $ch = curl_init();
- curl_setopt ($ch, CURLOPT_URL, "http://www.phpfensi.com/");
- curl_setopt ($ch, CURLOPT_REFERER, "http://www.phpfensi.com/");
- curl_exec ($ch);
- curl_close ($ch);
-
-
- $server = 'blog.qita.in';
- $host = 'blog.qita.in';
- $target = '/xxx.asp';
- $referer = 'http://www.baidu.com/'; // Referer
- $port = 80;
- $fp = fsockopen($server, $port, $errno, $errstr, 30);
- if (!$fp)
- {
- echo "$errstr ($errno)<br />n";
- }
- else
- {
- $out = "GET $target HTTP/1.1rn";
- $out .= "Host: $hostrn";
- $out .= "Cookie: ASPSESSIONIDSQTBQSDA=DFCAPKLBBFICDAFMHNKIGKEGrn";
- $out .= "Referer: $refererrn";
- $out .= "Connection: Closernrn";
- fwrite($fp, $out);
- while (!feof($fp))
- {
- echo fgets($fp, 128);
- }
- fclose($fp);
- }