使用 PhP 获取网络客户端 MAC 地址、语法问题
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Getting Network Client MAC address, Syntax issue using PhP
提问by Vijay
Actually the below coding is working fine, if I provide the ip address directly inside the shell_exec()
实际上下面的编码工作正常，如果我直接在里面提供IP地址 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,
如果，我从他的系统中获取客户端的 ip 并存储在一个变量中并调用它，如下所示，
$mac = shell_exec('arp -a' . escapeshellarg($ip));
The output is not generating.
输出没有生成。
Here is the Full code:
这是完整的代码：
<?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().
请指教，为什么escapeshellarg($ip)在shell_exec().
回答by cyber8200
shell_exec('arp '.$ip.' | awk \'{print $4}\'');
shell_exec('arp '.$ip.' | awk \'{print $4}\'');
Result from Terminal
结果来自终端
└── arp 10.1.10.26 | awk '{print $4}'
└── arp 10.1.10.26 | awk '{print $4}'
a4:5e:60:ee:29:19
a4:5e:60:ee:29:19
回答by Krumple
This is the correct format:
这是正确的格式：
$mac=shell_exec("arp -a ".$ip);
or
或者
$mac=shell_exec("arp -a ".escapeshellarg($ip));
(using the escapeshellargcall)
（使用escapeshellarg调用）
回答by roycruse
This is working for me...
这对我有用...
$ip=$_SERVER['REMOTE_ADDR'];
$mac_string = shell_exec("arp -a $ip");
$mac_array = explode(" ",$mac_string);
$mac = $mac_array[3];
echo($ip." - ".$mac);
回答by syl.fabre
A space is missing just after the -a in 'arp -a'.escape...
'arp -a'.escape 中的 -a 之后缺少一个空格...
So it turns into arp -a192.168.0.107
所以它变成了 arp -a192.168.0.107
回答by Vinay A
shell_exec("arp -a ".escapeshellarg($_SERVER['REMOTE_ADDR'])." | grep -o -E '([[:xdigit:]]{1,2}:){5}[[:xdigit:]]{1,2}'");
