How should I call this function? PHP

Question

How should I call this function? I'm new to PHP. Here's my code... but I have an error

Notice: Undefined variable: ip in C:\xampp\htdocs\PHPTest\ip.php on line 19

    <?php
    function getRealIpAddr(){
        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 $ip;
    }
    call_user_func('getRealIpAddr', '$ip');
    echo $ip;
    ?>

UPDATE

Strange reason, I'm using Windows 10, localhost, xampp and google Chrome this script doesn't provide me an ip address! That's why a corrected code was empty... Thought it was some kind of errors or something

Second UPDATE

If you're getting no ip like me, you may try this solution on httpd.conf

for update: read this: http://stackoverflow.com/questions/3699454/should-a-mamp-return-1-as-ip-on-localhost — devpro, Jan 28 '16 at 15:51
are u getting this: ::1 ??? and want to get this 127.0.0.1 ?? for UPDATE — devpro, Jan 28 '16 at 15:55
@devpro empty and just updated my httpd.conf to `Listen 127.0.0.1:80`, restarted still didn't work :o — DagicCross, Jan 28 '16 at 15:56
this question will help u: http://stackoverflow.com/questions/10982277/serverremote-addr-not-giving-the-right-ip-address — devpro, Jan 28 '16 at 15:57
More likely the issue is `call_user_func('getRealIpAddr', '$ip');` remove the single quotes around `'$ip'` Plus there is no parameter defined on that function to receive `$ip` anyway and it would need to be passed by reference the way you are doing it! — RiggsFolly, Jan 28 '16 at 15:57

score 4 · Answer 1 · answered Jan 28 '16 at 15:42

4

Your function returns the IP address, so assign a variable to the return value of the function, like so:

$ip = getRealIpAddr();

answered Jan 28 '16 at 15:42

mikeb

10,578
7
62
120

score 2 · Accepted Answer · answered Jan 28 '16 at 15:43

Error is very clear $ip is not defined:

Modified Code:

<?
function getRealIpAddr(){
    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 = getRealIpAddr(); // your function
echo $ip;
?>

If you want to use $ip variable that you defined inside the function than note that scope of the $ip is limited into the function.

You can not call this variable outside the function. for this you need to store it in a variable as like above mentioned example ($ip = getRealIpAddr();).

Thomas Rollet · Answer 3 · 2016-01-28T15:49:09.087

-1

You can't access to a variable declared inside a function, but with the return statement you can do :

echo getRealIpAddr();

Instead of :

call_user_func('getRealIpAddr', '$ip');
echo $ip;

edited Jan 28 '16 at 15:49

answered Jan 28 '16 at 15:42

Thomas Rollet

1,573
4
19
33

How should I call this function? PHP

3 Answers3