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Hello,

 

I am trying to detemine the MAC address of a machine on a local network from the IP address.

The following code does exactly that.

 

<?php
$IP="192.168.1.18";
exec("ping -c 1 -s 1 $IP");
$mac=exec("/usr/sbin/arp -an | /bin/grep $IP | /usr/bin/awk '{print $4}'");
echo "$IP - $mac";
?>

 

However I need to take the IP address from an HTML form. When using the $_SERVER['REMOTE_ADDR'] variable seems to screw things up for me. Maybe its not the correct type of variable? For example the following will not work.

 

<?php
$IP=$_SERVER['REMOTE_ADDR'];
exec("ping -c 1 -s 1 $IP");
$mac=exec("/usr/sbin/arp -an | /bin/grep $IP | /usr/bin/awk '{print $4}'");
echo "$IP - $mac";
?>

 

Can anyone explain why this is? Our suggest an alternative method of doing this?

 

Regards,

Colin

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https://forums.phpfreaks.com/topic/273319-_serverremote_addr-help/
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Echo out the variable, preferably using var_dump (), and see if it is what you expect it to be.

 

Also, do take note of the fact that the IP is provided by the client. Which means that you have to validate it before using it, as an attacker could quite simply spoof it.

Gave that a shot with...

 

<?php
#$IP="192.168.1.18";
$IP=$_SERVER['REMOTE_ADDR'];
var_dump($IP);
?><br /><?php
exec("ping -c 1 -s 1 $IP");
$ans=exec("/usr/sbin/arp -an | /bin/grep $IP | /usr/bin/awk '{print $4}'");
echo "$IP - $ans";
?>

 

and got

 

string(12) "192.168.1.30"
192.168.1.30 -

 

Still no idea why I can't pass it to exec correctly. I'm just trying to get the basic mechanism of getting the mac address in place. Will validate any input at a later stage though.

I think, the execution path to the arp table is wrong.

 

Does ip address 192.168.1.18 belong to the local machine?

 

There is nothing wrong about $_SERVER['REMOTE_ADDR'].

 

Try to change

 

/usr/sbin/arp -an

 

to

 

/sbin/arp -an

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