重新编译后如何获取元素？

Question

我使用“re”编译握手的数据，如下所示：

 piece_request_handshake = re.compile('13426974546f7272656e742070726f746f636f6c(?P<reserved>\w{16})(?P<info_hash>\w{40})(?P<peer_id>\w{40})')

 handshake = piece_request_handshake.findall(hex_data)

然后我打印出来

我无法添加图片，因为我是新手，所以这是输出：

root@debian:/home/florian/Téléchargements# python script.py 
[('0000000000100005', '606d4759c464c8fd0d4a5d8fc7a223ed70d31d7b', '2d5452323532302d746d6e6a657a307a6d687932')]

我的问题是，我怎样才能只获取该数据的第二部分，即“hash_info”（“606d47...”）？

我已经尝试过使用以下行的 re 组：

   print handshake.group('info_hash')

但结果是错误（再次抱歉我无法显示屏幕...）：

*root@debian:/home/florian/Téléchargements# python script.py 
Exception in thread Thread-1:
Traceback (most recent call last):
  File "/usr/lib/python2.7/threading.py", line 552, in __bootstrap_inner
    self.run()
  File "script.py", line 122, in run
    self.p.dispatch(0, PieceRequestSniffer.cb)
  File "script.py", line 82, in cb
    print handshake.group('info_hash')
AttributeError: 'list' object has no attribute 'group'*

这是我为好奇者准备的完整代码的开始：

import pcapy
import dpkt
from threading import Thread
import re
import binascii
import socket
import time

liste=[]
prefix = '13426974546f7272656e742070726f746f636f6c'
hash_code = re.compile('%s(?P<reserved>\w{16})(?P<info_hash>\w{40})(?P<peer_id>\w{40})' % prefix)
match = hash_code.match()
piece_request_handshake = re.compile('13426974546f7272656e742070726f746f636f6c(?P<aaa>\w{16})(?P<bbb>\w{40})(?P<ccc>\w{40})')
piece_request_tcpclose = re.compile('(?P<start>\w{12})5011')


#-----------------------------------------------------------------INIT------------------------------------------------------------

class PieceRequestSniffer(Thread):
    def __init__(self, dev='eth0'):
        Thread.__init__(self)

        self.expr = 'udp or tcp'

        self.maxlen = 65535  # max size of packet to capture
        self.promiscuous = 1  # promiscuous mode?
        self.read_timeout = 100  # in milliseconds
        self.max_pkts = -1  # number of packets to capture; -1 => no limit

        self.active = True
        self.p = pcapy.open_live(dev, self.maxlen, self.promiscuous, self.read_timeout)
        self.p.setfilter(self.expr)

    @staticmethod
    def cb(hdr, data):

        eth = dpkt.ethernet.Ethernet(str(data))
        ip = eth.data



#------------------------------------------------------IPV4 AND TCP PACKETS ONLY---------------------------------------------------           



            #Select Ipv4 packets because of problem with the .p in Ipv6
        if eth.type == dpkt.ethernet.ETH_TYPE_IP6:
            return
        else:

            #Select only TCP protocols
            if ip.p == dpkt.ip.IP_PROTO_TCP:
                tcp = ip.data

                src_ip = socket.inet_ntoa(ip.src)
                dst_ip = socket.inet_ntoa(ip.dst)

                fin_flag = ( tcp.flags & dpkt.tcp.TH_FIN ) != 0
                #if fin_flag:
                    #print "TH_FIN src:%s dst:%s" % (src_ip,dst_ip)




                try:
                    #Return hexadecimal representation
                    hex_data = binascii.hexlify(tcp.data)
                except:
                    return



#-----------------------------------------------------------HANDSHAKE-------------------------------------------------------------




                handshake = piece_request_handshake.findall(hex_data)
                if handshake and (src_ip+" "+dst_ip) not in liste and (dst_ip+" "+src_ip) not in liste and handshake != '':
                    liste.append(src_ip+" "+dst_ip)
                    print match.group('info_hash')

Answer 1

re.findall() 返回一个元组列表，每个元组都包含与 re 模式中的命名组相对应的匹配字符串。此示例（使用简化模式）演示了您可以通过索引访问所需的项目：

import re

prefix = 'prefix'
pattern = re.compile('%s(?P<reserved>\w{4})(?P<info_hash>\w{10})(?P<peer_id>\w{10})' % prefix)
handshake = 'prefix12341234567890ABCDEF1234'    # sniffed data
match = pattern.findall(handshake)

>>> print match
[('1234', '1234567890', 'ABCDEF1234')]
>>> info_hash = match[0][1]
>>> print info_hash
1234567890

但命名组的意义在于提供一种按名称访问命名组的匹配值的方法。您可以改用re.match()：

import re

prefix = 'prefix'
pattern = re.compile('%s(?P<reserved>\w{4})(?P<info_hash>\w{10})(?P<peer_id>\w{10})' % prefix)
handshake = 'prefix12341234567890ABCDEF1234'    # sniffed data
match = pattern.match(handshake)

>>> print match
<_sre.SRE_Match object at 0x7fc201efe918>
>>> print match.group('reserved')
1234
>>> print match.group('info_hash')
1234567890
>>> print match.group('peer_id')
ABCDEF1234

这些值也可以通过字典访问获得：

>>> d = match.groupdict()
>>> d
{'peer_id': 'ABCDEF1234', 'reserved': '1234', 'info_hash': '1234567890'}
>>> d['info_hash']
'1234567890'

最后，如果输入数据中有多个握手序列，可以使用re.finditer()：

import re

prefix = 'prefix'
pattern = re.compile('%s(?P<reserved>\w{4})(?P<info_hash>\w{10})(?P<peer_id>\w{10})' % prefix)
handshake = 'blahprefix12341234567890ABCDEF1234|randomjunkprefix12349876543210ABCDEF1234,more random junkprefix1234hellothereABCDEF1234...'    # sniffed data

for match in pattern.finditer(handshake):
    print match.group('info_hash')

输出：

1234567890 9876543210 你好

Answer 2

re.findall 将返回一个元组列表。 group() 调用适用于 Match 对象，由 re 中的一些其他函数返回：

for match in re.finditer(needle, haystack):
    print match.group('info_hash')

此外，如果您只是匹配一次握手，则可能不需要 findall。