Scapy 不接受这种值“10.0.2.1/24”

Question

我的代码应该发送数据包并取回 mac 地址，但我遇到了一些错误，代码：

def scan(ip):
    arp_request = scapy.ARP(pdst=ip)
    broadcast = scapy.Ether(dst="ff:ff:ff:ff:ff:ff")
    arp_request_broadcast = broadcast/arp_request
    print(arp_request_broadcast.summary())

scan("10.0.2.1/24")

错误是：

Ether / ARP who has ?? says ??

Answer 1

你传给ARP层的值就好了。但是你的代码没有sending/receiving任何数据包到网络。

让我们修复您的代码：


from scapy.all import ARP, Ether, srp
import scapy.all as scapy

def scan(ip):
    arp_request = scapy.ARP(pdst=ip)
    broadcast = scapy.Ether(dst="ff:ff:ff:ff:ff:ff")
    arp_request_broadcast = broadcast/arp_request

    # send packet to the network, wait for replies for 3 secondes, then return all the answers
    result = srp(arp_request_broadcast, timeout=3, verbose=0)[0]
    
    # for each replay, 
    for sent, received in result:
        print(received.summary())

scan("192.168.1.1/24")

在我的网络中给我：

Ether / ARP is at 00:24:46:03:87:11 says 192.168.1.195 / Padding
Ether / ARP is at f8:75:a4:76:68:36 says 192.168.1.216 / Padding
Ether / ARP is at 78:96:84:9c:13:20 says 192.168.1.254 / Padding

请注意，您需要运行 sudo 中的代码，因为它需要 read/write 到网络

你可以在这里看到完整的解决方案：https://www.thepythoncode.com/article/building-network-scanner-using-scapy

Scapy 不接受这种值“10.0.2.1/24”

The Scapy is not accepting this kind of value "10.0.2.1/24"

python

scapy

python-3.x

kali-linux