Every science begins as philosophy and ends as art….

function getIP(){
if (isset($_SERVER['HTTP_X_FORWARDED_FOR']) && eregi(“^[0-9]{1,3}\\.[0-9]{1,3}\\.[0-9]{1,3}\\.[0-9]{1,3}$”,$_SERVER['HTTP_X_FORWARDED_FOR'])){
$ip = $_SERVER['HTTP_X_FORWARDED_FOR'];
} else {
$ip = getenv(“REMOTE_ADDR”);
}
return $ip;
}

My problem with $_SERVER['remote_addr']

I’m not sure if I’m having a problem with the installation of PHP on the server or if there’s something that I’m not doing right in the code, but I am trying to capture the IP address of anyone that is trying to log into a secure part of a site. Regardless of my IP address, when I do:

Code:

echo $_SERVER['remote_addr'];

It returns ’192.168.1.1′. This is the first time I’ve written code that is being used on a Windows Server, so I don’t know if that has anything to do with it.

I’ve run phpinfo(), and in the PHP Variables section, it shows “_SERVER["REMOTE_ADDR"]” : 192.168.1.1. Does this mean that the php.ini file is setup wrong, or am I missing something?

i am running my own server on windows 7 iis7.5 with php as fastcgi and i am behind a router with a dynamic ip not static and i am using
http://www.no-ip.com/ that give me a link to my router and i am opening 80 port in my router forwarding it to my id address 192.168.1.5 and i have tested a php file that contains

tested it using tor network that give me ip out of my network to test that file always give me the same result 192.168.1.1 i am trying to get IP address of the visitor and i asked someone of my friends to use this form it sends the same ip

i hope someone tells me what i am doing wrong and the solution for this problem

Thanks in advance