CODEWITHSUNDEEP
phpip
imez
imez

Reputation: 308

problem with $_SERVER['REMOTE_ADDR']

i used $_SERVER['REMOTE_ADDR'] and it returns client ip address (IP address from which the user is viewing the current page) but at now (and same code) it returns host ip address (i checked ip address with ip location). problem is with host or what? thank u.

Upvotes: 1

Views: 25022

Answers (3)

Mark Bekkers
Mark Bekkers

Reputation: 437

@James @imez

By default the client IP is in $_SERVER['REMOTE_ADDR']. When the user enters your site using a PROXY server (HTTP gateway) it tells you who it's proxing for (HTTP_X_FORWARDED_FOR) and will give it's own Proxy IP in $_SERVER['REMOTE_ADDR'].

Anonymous proxies will omit HTTP_X_FORWARDED_FOR or simply lie to you.

Knowing you have a real client IP isn't possible.

Upvotes: 4

user7675
user7675

Reputation:

I have to mention that the array key is case-sensitive, and should be upper-case:

var_dump($_SERVER['remote_addr']);
echo "\n";
var_dump($_SERVER['REMOTE_ADDR']);

Output:

Notice: Undefined index: remote_addr in /home/adam/public_html/2011/01/04/foo.php on line 3
NULL

string(15) "10.0.1.51"

I would var_dump($_SERVER) just to evaluate the state of your world, and go from there.

Upvotes: 2

James
James

Reputation: 82096

You should query for HTTP_X_FORWARDED_FOR first and if it isn't assigned use REMOTE_ADDR.

Upvotes: 12

Related Questions