如何指定一个scapy启动文件？

Question

I use scapy with the -c command line option to load a startup file:

# liquidsoap debug
streamerIP = "192.168.0.53"
dump= []

def filterStreamer(pkt):
    if pkt.src == streamerIP or pkt.dst == streamerIP:
        dump.append(pkt)

sniff(prn=filterStreamer)

ls(dump)

it gives:

Traceback (most recent call last):
  File "/usr/lib/python2.7/dist-packages/scapy/main.py", line 30, in _read_config_file
    execfile(cf)
  File "icecast-debug.py", line 9, in <module>
    sniff(prn=filterStreamer)
  File "/usr/lib/python2.7/dist-packages/scapy/sendrecv.py", line 586, in sniff
    r = prn(p)
  File "icecast-debug.py", line 6, in filterStreamer
    if (pkt.src == streamerIP or pkt.dst == streamerIP):
NameError: global name 'streamerIP' is not defined
Welcome to Scapy (2.2.0)

and in the console I see nor the streamerIP neither dump, but funniest of all filterStreamer as a function are not defined. However if I do not pass filterStreamer to sniff it begins sniffing. So it's like interpreting the code line by line, and clear the scope after all line interpretition.

Answer 1

You have to use the global keyword. Also, use a PacketList() rather than a list. And ls() will not work against a list, but if you use a PacketList() , you have the .summary() method.

streamerIP = "192.168.0.53"
dump = PacketList()

def filterStreamer(pkt):
    global streamerIP, dump
    if pkt.src == streamerIP or pkt.dst == streamerIP:
        dump.append(pkt)

dump.summary()