检索准确的外部IPV4地址PHP

Question

I was watching a video on how to gather the real external IP address of the user in [YT: DewClarke ]'s video. His code worked perfectly on his website , but resulted with an IPV6 version on mine : 2701:0564:d210:5be4:5b0b:8630:4c99:a3be (randomized for privacy reasons). How can I get the IPV4 like 216.3.128.12 (randomized for privacy reasons)?

<?php 
    function getUserIP(){
      $client  = @$_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();
?>

I "echo" the data later down the line...

<div id="LoggedAs"><span><?php echo $user_ip;?></span></div>

Some useful information is that I use 000WebHost.com, run it through CloudFlare.com, it's written in PHP, and I force SSL on that page. DewClarke's source-code can be found here: DewClarke's Source Code

Answer 1

Your code seems fine. This seems like a Cloudflare setting. Go to Cloudflare -> (your domain) -> Network. Then set IPv6 Compatibility to OFF.

It is recommended though that you keep this setting to ON because IPv6 is critical to the growth of worldwide connection of networks.

You may also be interested to use this code instead for getting the IP address and country of the connecting user as provided by Cloudflare:

$_SERVER['REMOTE_ADDR'] = isset($_SERVER['HTTP_CF_CONNECTING_IP']) ? $_SERVER['HTTP_CF_CONNECTING_IP'] : $_SERVER['REMOTE_ADDR'];
$_SERVER['REMOTE_COUNTRY'] = isset($_SERVER['HTTP_CF_IPCOUNTRY']) ? $_SERVER['HTTP_CF_IPCOUNTRY'] : 'Unknown';