PHP curl偽造IP地址和header信息代碼實例

curl雖然功能強大，但是只能偽造$_SERVER["HTTP_X_FORWARDED_FOR"]，對於大多數IP地址檢測程序來說，$_SERVER["REMOTE_ADDR"]很難被偽造：

首先是client.php的代碼

復制代碼 代碼如下:
$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 , $headerArr );  //構造IP
curl_setopt ($ch, CURLOPT_REFERER, "http://www.163.com/ ");   //構造來路
curl_setopt( $ch, CURLOPT_HEADER, 1);
 
curl_exec($ch);
curl_close ($ch);
$out = ob_get_contents();
ob_clean();
 
echo $out;

然後是server.php

復制代碼 代碼如下:
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 = "無法獲取！";
    return $cip;
}
echo "<br>訪問IP: ".GetIP()."<br>";
echo "<br>訪問來路: ".$_SERVER["HTTP_REFERER"];