I need to create a simple webpage that only shows someones public IPV4 address in their browser window

Question

What I am looking to do is create a simple page, maybe in php, that only shows the IPV4 Address.

It should look like this site "http://myexternalip.com/raw" but not show IPV6 only IPV4.

This site shows my IPV6 address, and I know that I have both working because Whatsmyip.com shows my IPv4.

Thanks for any help everyone!

I would like to thank the people who are willing to give help answer questions, and for the people giving down reviews, why? I don't give a down review and then not leave a comment, that's just spamming really. — Ajster1989, Sep 16 '14 at 01:50

score 0 · Answer 1 · answered Sep 15 '14 at 02:23

The simplest way to get the visitor’s/client’s IP address is using the $_SERVER['REMOTE_ADDR'] or $_SERVER['REMOTE_HOST'] variables.

However, sometimes this does not return the correct IP address of the visitor, so we can use some other server variables to get the IP address.

The below both functions are equivalent with the difference only in how and from where the values are retrieved.

getenv() is used to get the value of an environment variable in PHP.

// Function to get the client IP address
function get_client_ip() {
    $ipaddress = '';
    if (getenv('HTTP_CLIENT_IP'))
        $ipaddress = getenv('HTTP_CLIENT_IP');
    else if(getenv('HTTP_X_FORWARDED_FOR'))
        $ipaddress = getenv('HTTP_X_FORWARDED_FOR');
    else if(getenv('HTTP_X_FORWARDED'))
        $ipaddress = getenv('HTTP_X_FORWARDED');
    else if(getenv('HTTP_FORWARDED_FOR'))
        $ipaddress = getenv('HTTP_FORWARDED_FOR');
    else if(getenv('HTTP_FORWARDED'))
        $ipaddress = getenv('HTTP_FORWARDED');
    else if(getenv('REMOTE_ADDR'))
        $ipaddress = getenv('REMOTE_ADDR');
    else
        $ipaddress = 'UNKNOWN';
    return $ipaddress;
}

$_SERVER is an array that contains server variables created by the web server.

// Function to get the client IP address
function get_client_ip() {
    $ipaddress = '';
    if ($_SERVER['HTTP_CLIENT_IP'])
        $ipaddress = $_SERVER['HTTP_CLIENT_IP'];
    else if($_SERVER['HTTP_X_FORWARDED_FOR'])
        $ipaddress = $_SERVER['HTTP_X_FORWARDED_FOR'];
    else if($_SERVER['HTTP_X_FORWARDED'])
        $ipaddress = $_SERVER['HTTP_X_FORWARDED'];
    else if($_SERVER['HTTP_FORWARDED_FOR'])
        $ipaddress = $_SERVER['HTTP_FORWARDED_FOR'];
    else if($_SERVER['HTTP_FORWARDED'])
        $ipaddress = $_SERVER['HTTP_FORWARDED'];
    else if($_SERVER['REMOTE_ADDR'])
        $ipaddress = $_SERVER['REMOTE_ADDR'];
    else
        $ipaddress = 'UNKNOWN';
    return $ipaddress;
}

Thanks for your help, I ended up going with the much shorter version, thanks anyways! — Ajster1989, Sep 16 '14 at 01:53

score 0 · Answer 2 · edited May 23 '17 at 12:29

0

Create a ip.php file and put the following code in it:

<?php
if (!empty($_SERVER['HTTP_CLIENT_IP'])) {
  $ip = $_SERVER['HTTP_CLIENT_IP'];
} elseif (!empty($_SERVER['HTTP_X_FORWARDED_FOR'])) {
  $ip = $_SERVER['HTTP_X_FORWARDED_FOR'];
} else {
  $ip = $_SERVER['REMOTE_ADDR'];
}
echo $ip;

My answer is based on this answer to a similar question: https://stackoverflow.com/a/55790/326016

edited May 23 '17 at 12:29

Community

1
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answered Sep 15 '14 at 02:23

Abenil

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This was the simple and perfect answer I was looking for, If I could I would hit that up arrow 10 times! Thanks! – Ajster1989 Sep 16 '14 at 01:52

I need to create a simple webpage that only shows someones public IPV4 address in their browser window

2 Answers2