我知道有大量的$_SERVER变量头可用于IP地址检索。我想知道是否有一个普遍的共识,如何最准确地检索用户的真实IP地址(好知道没有方法是完美的)使用上述变量?

我花了一些时间试图找到一个深入的解决方案,并根据一些来源提出了以下代码。如果有人能在答案中找出漏洞,或者提供一些更准确的信息,我会很高兴。

edit包含来自@Alix的优化

 /**
  * Retrieves the best guess of the client's actual IP address.
  * Takes into account numerous HTTP proxy headers due to variations
  * in how different ISPs handle IP addresses in headers between hops.
  */
 public function get_ip_address() {
  // Check for shared internet/ISP IP
  if (!empty($_SERVER['HTTP_CLIENT_IP']) && $this->validate_ip($_SERVER['HTTP_CLIENT_IP']))
   return $_SERVER['HTTP_CLIENT_IP'];

  // Check for IPs passing through proxies
  if (!empty($_SERVER['HTTP_X_FORWARDED_FOR'])) {
   // Check if multiple IP addresses exist in var
    $iplist = explode(',', $_SERVER['HTTP_X_FORWARDED_FOR']);
    foreach ($iplist as $ip) {
     if ($this->validate_ip($ip))
      return $ip;
    }
   }
  }
  if (!empty($_SERVER['HTTP_X_FORWARDED']) && $this->validate_ip($_SERVER['HTTP_X_FORWARDED']))
   return $_SERVER['HTTP_X_FORWARDED'];
  if (!empty($_SERVER['HTTP_X_CLUSTER_CLIENT_IP']) && $this->validate_ip($_SERVER['HTTP_X_CLUSTER_CLIENT_IP']))
   return $_SERVER['HTTP_X_CLUSTER_CLIENT_IP'];
  if (!empty($_SERVER['HTTP_FORWARDED_FOR']) && $this->validate_ip($_SERVER['HTTP_FORWARDED_FOR']))
   return $_SERVER['HTTP_FORWARDED_FOR'];
  if (!empty($_SERVER['HTTP_FORWARDED']) && $this->validate_ip($_SERVER['HTTP_FORWARDED']))
   return $_SERVER['HTTP_FORWARDED'];

  // Return unreliable IP address since all else failed
  return $_SERVER['REMOTE_ADDR'];
 }

 /**
  * Ensures an IP address is both a valid IP address and does not fall within
  * a private network range.
  *
  * @access public
  * @param string $ip
  */
 public function validate_ip($ip) {
     if (filter_var($ip, FILTER_VALIDATE_IP, 
                         FILTER_FLAG_IPV4 | 
                         FILTER_FLAG_IPV6 |
                         FILTER_FLAG_NO_PRIV_RANGE | 
                         FILTER_FLAG_NO_RES_RANGE) === false)
         return false;
     self::$ip = $ip;
     return true;
 }

警告之词(更新)

REMOTE_ADDR仍然表示最可靠的IP地址来源。这里提到的其他$_SERVER变量很容易被远程客户端欺骗。此解决方案的目的是试图确定位于代理后面的客户机的IP地址。出于一般目的,您可以考虑将其与直接从$_SERVER['REMOTE_ADDR']返回的IP地址结合使用,并存储两者。

对于99.9%的用户,这个解决方案将完美地满足您的需求。它不能保护您免受0.1%的恶意用户通过注入他们自己的请求头来滥用您的系统。如果某些关键任务依赖于IP地址,请使用REMOTE_ADDR,而不必费心满足代理背后的需求。


当前回答

我很惊讶没有人提到filter_input,所以这里是Alix Axel的回答浓缩成一行:

function get_ip_address(&$keys = ['HTTP_X_FORWARDED_FOR', 'HTTP_X_FORWARDED', 'HTTP_X_CLUSTER_CLIENT_IP', 'HTTP_FORWARDED_FOR', 'HTTP_FORWARDED', 'HTTP_CLIENT_IP', 'REMOTE_ADDR'])
{
    return empty($keys) || ($ip = filter_input(INPUT_SERVER, array_pop($keys), FILTER_VALIDATE_IP, FILTER_FLAG_NO_PRIV_RANGE | FILTER_FLAG_NO_RES_RANGE))? $ip : get_ip_address($keys);
}

其他回答

我们使用:

/**
 * Get the customer's IP address.
 *
 * @return string
 */
public function getIpAddress() {
    if (!empty($_SERVER['HTTP_CLIENT_IP'])) {
        return $_SERVER['HTTP_CLIENT_IP'];
    } else if (!empty($_SERVER['HTTP_X_FORWARDED_FOR'])) {
        $ips = explode(',', $_SERVER['HTTP_X_FORWARDED_FOR']);
        return trim($ips[count($ips) - 1]);
    } else {
        return $_SERVER['REMOTE_ADDR'];
    }
}

在HTTP_X_FORWARDED_FOR上的爆炸是因为我们在使用Squid时检测IP地址的奇怪问题。

/**
 * Sanitizes IPv4 address according to Ilia Alshanetsky's book
 * "php|architect?s Guide to PHP Security", chapter 2, page 67.
 *
 * @param string $ip An IPv4 address
 */
public static function sanitizeIpAddress($ip = '')
{
if ($ip == '')
    {
    $rtnStr = '0.0.0.0';
    }
else
    {
    $rtnStr = long2ip(ip2long($ip));
    }

return $rtnStr;
}

//---------------------------------------------------

/**
 * Returns the sanitized HTTP_X_FORWARDED_FOR server variable.
 *
 */
public static function getXForwardedFor()
{
if (isset($_SERVER['HTTP_X_FORWARDED_FOR']))
    {
    $rtnStr = $_SERVER['HTTP_X_FORWARDED_FOR'];
    }
elseif (isset($HTTP_SERVER_VARS['HTTP_X_FORWARDED_FOR']))
    {
    $rtnStr = $HTTP_SERVER_VARS['HTTP_X_FORWARDED_FOR'];
    }
elseif (getenv('HTTP_X_FORWARDED_FOR'))
    {
    $rtnStr = getenv('HTTP_X_FORWARDED_FOR');
    }
else
    {
    $rtnStr = '';
    }

// Sanitize IPv4 address (Ilia Alshanetsky):
if ($rtnStr != '')
    {
    $rtnStr = explode(', ', $rtnStr);
    $rtnStr = self::sanitizeIpAddress($rtnStr[0]);
    }

return $rtnStr;
}

//---------------------------------------------------

/**
 * Returns the sanitized REMOTE_ADDR server variable.
 *
 */
public static function getRemoteAddr()
{
if (isset($_SERVER['REMOTE_ADDR']))
    {
    $rtnStr = $_SERVER['REMOTE_ADDR'];
    }
elseif (isset($HTTP_SERVER_VARS['REMOTE_ADDR']))
    {
    $rtnStr = $HTTP_SERVER_VARS['REMOTE_ADDR'];
    }
elseif (getenv('REMOTE_ADDR'))
    {
    $rtnStr = getenv('REMOTE_ADDR');
    }
else
    {
    $rtnStr = '';
    }

// Sanitize IPv4 address (Ilia Alshanetsky):
if ($rtnStr != '')
    {
    $rtnStr = explode(', ', $rtnStr);
    $rtnStr = self::sanitizeIpAddress($rtnStr[0]);
    }

return $rtnStr;
}

//---------------------------------------------------

/**
 * Returns the sanitized remote user and proxy IP addresses.
 *
 */
public static function getIpAndProxy()
{
$xForwarded = self::getXForwardedFor();
$remoteAddr = self::getRemoteAddr();

if ($xForwarded != '')
    {
    $ip    = $xForwarded;
    $proxy = $remoteAddr;
    }
else
    {
    $ip    = $remoteAddr;
    $proxy = '';
    }

return array($ip, $proxy);
}

最大的问题是目的是什么?

你的代码几乎是全面的,因为它可以-但我看到,如果你发现什么看起来像一个代理添加头,你使用它而不是CLIENT_IP,然而,如果你想要这个信息的审计目的,然后警告-它很容易伪造。

当然,您绝不应该使用IP地址进行任何形式的身份验证——即使这些地址也可能被欺骗。

您可以通过推出一个flash或java applet来更好地测量客户端ip地址,它通过非http端口连接回服务器(因此会显示透明代理或代理注入的报头为假的情况下),但请记住,当客户端只能通过web代理连接或传出端口被阻塞时,将不会有来自applet的连接。

你已经回答了你自己的问题!:)

function getRealIpAddr() {
    if(!empty($_SERVER['HTTP_CLIENT_IP']))   //Check IP address from shared Internet
    {
        $IPaddress = $_SERVER['HTTP_CLIENT_IP'];
    }
    elseif (!empty($_SERVER['HTTP_X_FORWARDED_FOR']))   //To check IP address is passed from the proxy
    {
        $IPaddress = $_SERVER['HTTP_X_FORWARDED_FOR'];
    }
    else
    {
        $IPaddress = $_SERVER['REMOTE_ADDR'];
    }
    return $IPaddress;
}

虽然这篇文章很老了,但这个话题仍然值得关注。所以我在我的项目中使用了另一个解决方案。我在这里找到了其他的解决方案,要么不完整,要么太复杂,难以理解。

if (! function_exists('get_visitor_IP'))
{
    /**
     * Get the real IP address from visitors proxy. e.g. Cloudflare
     *
     * @return string IP
     */
    function get_visitor_IP()
    {
        // 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"];
        }

        // Sometimes the `HTTP_CLIENT_IP` can be used by proxy servers
        $ip = @$_SERVER['HTTP_CLIENT_IP'];
        if (filter_var($ip, FILTER_VALIDATE_IP)) {
           return $ip;
        }

        // Sometimes the `HTTP_X_FORWARDED_FOR` can contain more than IPs 
        $forward_ips = @$_SERVER['HTTP_X_FORWARDED_FOR'];
        if ($forward_ips) {
            $all_ips = explode(',', $forward_ips);

            foreach ($all_ips as $ip) {
                if (filter_var($ip, FILTER_VALIDATE_IP, FILTER_FLAG_NO_PRIV_RANGE | FILTER_FLAG_NO_RES_RANGE)){
                    return $ip;
                }
            }
        }

        return $_SERVER['REMOTE_ADDR'];
    }
}