103

我正在使用这个 PHP 代码来获取访问者的 IP 地址：

<?php echo $_SERVER['REMOTE_ADDR']; ?>

但是，当他们使用代理时，我无法从访问者那里获得真实的 IP 地址。在这种情况下，有什么方法可以获取访问者的 IP 地址吗？

4

9 回答 9

194

试试这个 php 代码。

<?PHP

function getUserIP()
{
    // Get real visitor IP behind CloudFlare network
    if (isset($_SERVER["HTTP_CF_CONNECTING_IP"])) {
              $_SERVER['REMOTE_ADDR'] = $_SERVER["HTTP_CF_CONNECTING_IP"];
              $_SERVER['HTTP_CLIENT_IP'] = $_SERVER["HTTP_CF_CONNECTING_IP"];
    }
    $client  = @$_SERVER['HTTP_CLIENT_IP'];
    $forward = @$_SERVER['HTTP_X_FORWARDED_FOR'];
    $remote  = $_SERVER['REMOTE_ADDR'];

    if(filter_var($client, FILTER_VALIDATE_IP))
    {
        $ip = $client;
    }
    elseif(filter_var($forward, FILTER_VALIDATE_IP))
    {
        $ip = $forward;
    }
    else
    {
        $ip = $remote;
    }

    return $ip;
}


$user_ip = getUserIP();

echo $user_ip; // Output IP address [Ex: 177.87.193.134]


?>
于 2012-11-30T14:01:13.630 回答
56

这是我见过的最常见的技术：

function getUserIP() {
    if( array_key_exists('HTTP_X_FORWARDED_FOR', $_SERVER) && !empty($_SERVER['HTTP_X_FORWARDED_FOR']) ) {
        if (strpos($_SERVER['HTTP_X_FORWARDED_FOR'], ',')>0) {
            $addr = explode(",",$_SERVER['HTTP_X_FORWARDED_FOR']);
            return trim($addr[0]);
        } else {
            return $_SERVER['HTTP_X_FORWARDED_FOR'];
        }
    }
    else {
        return $_SERVER['REMOTE_ADDR'];
    }
}

请注意，它并不能保证您将始终获得正确的用户 IP，因为有很多方法可以隐藏它。

于 2012-11-30T14:08:55.413 回答
16

这是我的方法：

 function getRealUserIp(){
    switch(true){
      case (!empty($_SERVER['HTTP_X_REAL_IP'])) : return $_SERVER['HTTP_X_REAL_IP'];
      case (!empty($_SERVER['HTTP_CLIENT_IP'])) : return $_SERVER['HTTP_CLIENT_IP'];
      case (!empty($_SERVER['HTTP_X_FORWARDED_FOR'])) : return $_SERVER['HTTP_X_FORWARDED_FOR'];
      default : return $_SERVER['REMOTE_ADDR'];
    }
 }

如何使用：

$ip = getRealUserIp();
于 2015-03-03T23:36:53.647 回答
6

代理可能会发送HTTP_X_FORWARDED_FOR标头，但即使是可选的。

还要记住，访问者可能会共享 IP 地址；大学网络、大公司和第三世界/低预算的 ISP 倾向于与许多用户共享 IP。

于 2012-11-30T13:59:33.903 回答
4

应用此代码获取 IP 地址：

    if (getenv('HTTP_X_FORWARDED_FOR')) { $pipaddress = getenv('HTTP_X_FORWARDED_FOR');
 $ipaddress = getenv('REMOTE_ADDR'); 
    echo "Your Proxy IP address is : ".$pipaddress. "(via $ipaddress)" ; } 
    else { $ipaddress = getenv('REMOTE_ADDR'); echo "Your IP address is : $ipaddress"; }
    ------------------------------------------------------------------------
于 2016-02-22T08:51:15.313 回答
4

This is my function.

benefits :

  • Work if $_SERVER was not available.
  • Filter private and/or reserved IPs;
  • Process all forwarded IPs in X_FORWARDED_FOR
  • Compatible with CloudFlare
  • Can set a default if no valid IP found!
  • Short & Simple !
/**
 * Get real user ip
 *
 * Usage sample:
 * GetRealUserIp();
 * GetRealUserIp('ERROR',FILTER_FLAG_NO_RES_RANGE);
 * 
 * @param string $default default return value if no valid ip found
 * @param int    $filter_options filter options. default is FILTER_FLAG_NO_PRIV_RANGE | FILTER_FLAG_NO_RES_RANGE
 *
 * @return string real user ip
 */

function GetRealUserIp($default = NULL, $filter_options = 12582912) {
    $HTTP_X_FORWARDED_FOR = isset($_SERVER)? $_SERVER["HTTP_X_FORWARDED_FOR"]:getenv('HTTP_X_FORWARDED_FOR');
    $HTTP_CLIENT_IP = isset($_SERVER)?$_SERVER["HTTP_CLIENT_IP"]:getenv('HTTP_CLIENT_IP');
    $HTTP_CF_CONNECTING_IP = isset($_SERVER)?$_SERVER["HTTP_CF_CONNECTING_IP"]:getenv('HTTP_CF_CONNECTING_IP');
    $REMOTE_ADDR = isset($_SERVER)?$_SERVER["REMOTE_ADDR"]:getenv('REMOTE_ADDR');

    $all_ips = explode(",", "$HTTP_X_FORWARDED_FOR,$HTTP_CLIENT_IP,$HTTP_CF_CONNECTING_IP,$REMOTE_ADDR");
    foreach ($all_ips as $ip) {
        if ($ip = filter_var($ip, FILTER_VALIDATE_IP, $filter_options))
            break;
    }
    return $ip?$ip:$default;
}
于 2017-05-02T12:47:33.680 回答
1

如果代理是您信任的，您可以尝试：（假设代理 IP 是151.101.2.10）

<?php

$trustProxyIPs = ['151.101.2.10'];

$clientIP  = isset($_SERVER['REMOTE_ADDR']) ? $_SERVER['REMOTE_ADDR'] : NULL;

if (in_array($clientIP, $trustProxyIPs)) {

    $headers = ['HTTP_CLIENT_IP', 'HTTP_X_FORWARDED_FOR'];

    foreach ($headers as $key => $header) {

        if (isset($_SERVER[$header]) && filter_var($_SERVER[$header], FILTER_VALIDATE_IP)) {

            $clientIP = $_SERVER[$header];

            break;
        }
    }
}

echo $clientIP;

这将防止直接请求的客户端伪造转发头，并通过受信任的代理获取真实 IP。

于 2017-07-08T02:15:55.940 回答
-1

是的，$_SERVER["HTTP_X_FORWARDED_FOR"]这是我在我的 nginx 服务器上的代理下查看我的 ip 的方式。

但是你最好的选择是phpinfo()在代理下请求的页面上运行，这样你就可以查看所有可用的变量并查看带有你真实 ip 的那个是什么。

于 2012-11-30T14:03:15.140 回答
-6

这适用于 Windows 和 Linux！不管是本地主机还是在线..

    function getIP() {
    $ip = $_SERVER['SERVER_ADDR'];

    if (PHP_OS == 'WINNT'){
        $ip = getHostByName(getHostName());
    }

    if (PHP_OS == 'Linux'){
        $command="/sbin/ifconfig";
        exec($command, $output);
        // var_dump($output);
        $pattern = '/inet addr:?([^ ]+)/';

        $ip = array();
        foreach ($output as $key => $subject) {
            $result = preg_match_all($pattern, $subject, $subpattern);
            if ($result == 1) {
                if ($subpattern[1][0] != "127.0.0.1")
                $ip = $subpattern[1][0];
            }
        //var_dump($subpattern);
        }
    }
    return $ip;
}
于 2015-11-17T13:45:37.640 回答