Getting Network Client MAC address, Syntax issue using PhP

Question

Actually the below coding is working fine, if I provide the ip address directly inside the shell_exec()

$mac = shell_exec('arp -a 192.168.0.107');

If, I get the ip of the client from his system and stored in a variable and call the same, as given below,

$mac = shell_exec('arp -a' . escapeshellarg($ip));

The output is not generating.

Here is the Full code:

<?php

$ip = $_SERVER['REMOTE_ADDR'];
$mac = shell_exec('arp -a'. escapeshellarg($ip));

//Working fine when sample client IP is provided...
//$mac = shell_exec('arp -a 192.168.0.107'); 

$findme = "Physical";
$pos = strpos($mac, $findme);
$macp = substr($mac,($pos+42),26);

if(empty($mac))
{
    die("No mac address for $ip not found");
}

// having it
echo "mac address for $ip: $macp";

?>

Please advise, why escapeshellarg($ip) does not work in the shell_exec().

Did you get your answer yet ? If not, I can help you. – code-8 Oct 09 '15 at 19:36 — code-8, Oct 09 '15 at 19:36

score 2 · Answer 1 · answered Oct 09 '15 at 19:35

2

shell_exec('arp '.$ip.' | awk \'{print $4}\'');

Result from Terminal

└── arp 10.1.10.26 | awk '{print $4}'

a4:5e:60:ee:29:19

answered Oct 09 '15 at 19:35

code-8

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Always makes me smile when people start correctly using awk commands! – Wes Mar 24 '16 at 19:37

score 1 · Answer 2 · edited Oct 11 '14 at 21:56

1

This is the correct format:

$mac=shell_exec("arp -a ".$ip);

or

$mac=shell_exec("arp -a ".escapeshellarg($ip));

(using the escapeshellarg call)

edited Oct 11 '14 at 21:56

Rick77

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answered Oct 11 '14 at 20:21

Krumple

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score 0 · Answer 3 · answered Jul 24 '14 at 08:55

0

A space is missing just after the -a in 'arp -a'.escape...

So it turns into arp -a192.168.0.107

answered Jul 24 '14 at 08:55

syl.fabre

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No I have removed and adding the space and tried several times, but it is not working. – Vijay Jul 24 '14 at 08:58
update your answer accordingly because in your code, a space is missing ! – syl.fabre Jul 24 '14 at 08:59
have you tried to do var_dump(escapeshellarg($ip)); ? – syl.fabre Jul 24 '14 at 08:59
if i use var_dump, It is listing all the ip and their mac addresses. – Vijay Jul 24 '14 at 09:02
I meant, use var_dump(escapeshellarg($ip)) to be sure that 'arp -a' . escapeshellarg($ip) == 'arp -a 192.168.0.107' – syl.fabre Jul 24 '14 at 09:05
actually i alter the coding like this ** $mac = shell_exec('arp -a' . var_dump(escapeshellarg($ip)));** it will list all the IP and Mac address in my networks – Vijay Jul 24 '14 at 09:08
use the var_dump outside the call to shell_exec. – syl.fabre Jul 24 '14 at 09:10
the purpose is to verify that 'arp -a' . escapeshellarg($ip) equals to 'arp -a 192.168.0.107' as you're claiming that $mac = shell_exec('arp -a 192.168.0.107'); works but $mac = shell_exec('arp -a' . escapeshellarg($ip)); doesn't – syl.fabre Jul 24 '14 at 09:11
$ip = var_dump(escapeshellarg($ip)); $mac = shell_exec('arp -a' . escapeshellarg($ip)); no use... $ip = var_dump(escapeshellarg($ip)); $mac = shell_exec('arp -a' . $ip); no use... till all the ip are listed – Vijay Jul 24 '14 at 09:14
so how can i call the variable $ip inside shell_exec()? – Vijay Jul 24 '14 at 09:19
do you know var_dump function ? – syl.fabre Jul 24 '14 at 09:20
Actually i stored the client system ip address in $ip " $ip = $_SERVER['REMOTE_ADDR'];" – Vijay Jul 24 '14 at 09:20
Actually i am new to php... i didn't know deeply. – Vijay Jul 24 '14 at 09:21
ok var_dump is like a debug function to display the content of a variable. It's not supposed to be used like $my_var = var_dump($something). You can use it to check that 'arp -a' . escapeshellarg($ip) is equals to 'arp -a 192.168.0.107' by doing on another line of code var_dump('arp -a' . escapeshellarg($ip)); to see if you're really executing 'arp -a 192.168.0.107' as you think you are – syl.fabre Jul 24 '14 at 09:25
Actually my need is to get the client ip and mac address who use my php program and stored in my database with time, employee name and employee no. I will now able to store client ip using $_SERVER['REMOTE_ADDR']; in DB. However I can't get the MAC address of the client system/person whom using my programs. – Vijay Jul 24 '14 at 09:38
So the getting ip value are related to MAc address. If i give the ip address in the position "escapeshellarg($ip)". I can getting my answer... If I use "escapeshellarg($ip)" the answer is not generating. It is not possible to assign IP as $mac = shell_exec('arp -a 192.168.0.107'); since it is a variable. – Vijay Jul 24 '14 at 09:43

score 0 · Answer 4 · answered Nov 19 '14 at 14:04

0

This is working for me...

$ip=$_SERVER['REMOTE_ADDR'];
$mac_string = shell_exec("arp -a $ip");
$mac_array = explode(" ",$mac_string);
$mac = $mac_array[3];
echo($ip." - ".$mac);

answered Nov 19 '14 at 14:04

roycruse

1
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score -1 · Answer 5 · edited Jul 29 '16 at 07:33

-1

shell_exec("arp -a ".escapeshellarg($_SERVER['REMOTE_ADDR'])." | grep -o -E '([[:xdigit:]]{1,2}:){5}[[:xdigit:]]{1,2}'");

edited Jul 29 '16 at 07:33

HiDeoo

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answered Jul 29 '16 at 06:11

Vinay A

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Getting Network Client MAC address, Syntax issue using PhP

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