PHP curl伪造IP地址和header信息代码实例_php实例_脚本之家
2015-05-01 10:08:49
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已经帮助：人解决问题
curl虽然功能强大，但是只能伪造$_SERVER["HTTP_X_FORWARDED_FOR"]，对于大多数IP地址检测程序来说，$_SERVER["REMOTE_ADDR"]很难被伪造：
首先是client.php的代码
复制代码 代码如下:
$headers['CLIENT-IP'] = '202.103.229.40';
$headers['X-FORWARDED-FOR'] = '202.103.229.40';
$headerArr = array();
foreach( $headers as $n => $v ) {
$headerArr[] = $n .':' . $v;
}
ob_start();
$ch = curl_init();
curl_setopt ($ch, CURLOPT_URL, "http://localhost/curl/server.php");
curl_setopt ($ch, CURLOPT_HTTPHEADER , $headerArr ); //构造IP
curl_setopt ($ch, CURLOPT_REFERER, "http://www.163.com/ "); //构造来路
curl_setopt( $ch, CURLOPT_HEADER, 1);
curl_exec($ch);
curl_close ($ch);
$out = ob_get_contents();
ob_clean();
echo $out;
然后是server.php
复制代码 代码如下:
function GetIP(){
if(!emptyempty($_SERVER["HTTP_CLIENT_IP"]))
$cip = $_SERVER["HTTP_CLIENT_IP"];
else if(!emptyempty($_SERVER["HTTP_X_FORWARDED_FOR"]))
$cip = $_SERVER["HTTP_X_FORWARDED_FOR"];
else if(!emptyempty($_SERVER["REMOTE_ADDR"]))
$cip = $_SERVER["REMOTE_ADDR"];
else
$cip = "无法获取！";
return $cip;
}
echo "<br>访问IP: ".GetIP()."<br>";
echo "<br>访问来路: ".$_SERVER["HTTP_REFERER"];
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