how to show The IP address in IPv4 which show ::1

问题:

I want to show the IP address of the computer's client. But in my computer which running in localhost show only "::1" . If i run in the localhost, it should be show 127.0.0.1. So how to show the IP address especially in IPv4. Because I read in another article that the ::1 is in IPv6. Here is my code :

function get_ip()
{
    if (!empty($_SERVER['HTTP_CLIENT_IP']))   //check ip from share internet
    {
      $ip=$_SERVER['HTTP_CLIENT_IP'];
    }
    elseif (!empty($_SERVER['HTTP_X_FORWARDED_FOR']))   //to check ip is pass from proxy
    {
      $ip=$_SERVER['HTTP_X_FORWARDED_FOR'];
    }
    else
    {
      $ip=$_SERVER['REMOTE_ADDR'];
    }
    return $ip;
}

$ip = get_ip(); 

echo $ip;

Give me help to fix this. Thank You.

回答1:

If you want the web server to see you connecting from 127.0.0.1 then you must connect via IPv4. Try navigating to http://127.0.0.1 instead of http://localhost. If you are connecting via IPv6 then of course the web server will report an IPv6 address.

回答2:

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

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

    if (PHP_OS == 'Linux'){
        $command="/sbin/ifconfig";
        exec($command, $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];
            }

        }
    }

    return $ip;
}

echo getIP();