GET 메서드로만 HTTP 패킷을 인쇄하는 Scapy에서 간단한 스니퍼를 코딩하려고합니다. 여기Scapy의 HTTP GET 패킷 스니퍼
#!/usr/bin/python
from scapy.all import *
def http_header(packet):
http_packet=str(packet)
if http_packet.find('GET'):
print GET_print(packet)
print packet
def GET_print(packet1):
print "***************************************GET PACKET****************************************************"
print packet1
print "*****************************************************************************************************"
sniff(iface='eth0',prn=http_header)
출력됩니다 :
*****************************************************************************************************
None
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***************************************GET PACKET****************************************************
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*****************************************************************************************************
내가 기대하고하는 것은 다음은 코드는
GET/HTTP/1.1
Host: google.com
User-Agent: Mozilla/5.0 (X11; Linux x86_64; rv:24.0) Gecko/20140722 Firefox/24.0 Iceweasel/24.7.0
Accept: text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8
Accept-Language: en-US,en;q=0.5
Accept-Encoding: gzip, deflate
Cookie: PREF=ID=758a20b5fbd4eac9:U=2b2dedf6c84b001f:FF=0:TM=1412150291:LM=1415430021:S=Q-QemmrLqsSsEA9i; NID=67=mRdkPVhtImrOTLi5I1e5JM22J7g26jAcdiDEjj9C5q0H5jj0DWRX27hCM7gLJBeiowW-8omSv-1ycH595SW2InWX2n1JMMNh6b6ZrRsZ9zOCC2a-vstOQnBDSJu6K9LO
Connection: keep-alive
내 예상 출력을 얻기 위해 무엇을 할 수 있는가?
감사! 매력처럼 작동합니다! –
당신의 대답은 아주 좋습니다. 필터링 작업을 수행하고, http_header()를 제거하고,'sniff (iface = 'eth0', prn = GET_print, lfilter = lambda p : str (p)의 "GET", 필터 = "tcp port 80")'. – Pierre
Thanks @Pierre 코드를 포함하도록 답변을 업데이트했습니다. –