我正试图从以下代码中获取2个ipv6地址:
import re
from pprint import pprint
with open('show_ipv6_intf.txt') as f:
show_ipv6_int = f.read()
match = re.search(r'^s+(S+)s[VALID]s+(S+).*', show_ipv6_int,flags=re.M)
ipv6_1 = match.group(1)
ipv6_2 = match.group(2)
ipv6_list = []
ipv6_list.append(ipv6_1)
ipv6_list.append(ipv6_2)
print(ipv6_list)
对于宽度,
这是正在使用的文件:
Ethernet2/4, Interface status: protocol-up/link-up/admin-up, iod: 40
IPv6 address:
2001:11:2233::a1/24 [VALID]
2001:cc11:22bb:0:2ec2:60ff:fe4f:feb2/64
IPv6 subnet: 2001::/24
IPv6 link-local address: fe80::2ec2:60ff:fe4f:feb2 (default) [VALID]
IPv6 virtual addresses configured: none
IPv6 multicast routing: disabled
IPv6 report link local: disabled
IPv6 Forwarding feature: disabled
IPv6 multicast groups locally joined:
ff02::1:ff4f:feb2 ff02::2 ff02::1 ff02::1:ff00:a1
ff02::1:ff4f:feb2 ff02::1:ff00:0
IPv6 multicast (S,G) entries joined: none
IPv6 MTU: 1500 (using link MTU)
IPv6 unicast reverse path forwarding: none
IPv6 load sharing: none
IPv6 interface statistics last reset: never
IPv6 interface RP-traffic statistics: (forwarded/originated/consumed)
Unicast packets: 0/0/0
Unicast bytes: 0/0/0
Multicast packets: 0/18/0
Multicast bytes: 0/2076/0
如果我使用标志";关于M";我检索期望的结果,但当我使用";关于"DOTALL";我将";无";布尔值。见下文:
带有re.M
C:UsersKenyone.PyCharm2019.3configscratches>python week4lesson6.py ['2001:11:2233::a1/24', '2001:cc11:22bb:0:2ec2:60ff:fe4f:feb2/64']
与re.DOTALL
C:UsersKenyone.PyCharm2019.3configscratches>python week4lesson6.py Traceback (most recent call last): File "week4lesson6.py", line 7, in <module> ipv6_1 = match.group(1) AttributeError: 'NoneType' object has no attribute 'group'
您希望从re.DOTALL获得什么?它改变了正则表达式模式中.
的行为,但您只有一个,没有必要。由于后面跟有*
,所以它可以匹配零次。你可以在没有点星的情况下重写模式,得到同样的行为。
r'^s+(S+)s[VALID]s+(S+)'
应该是相同的并且没有点。
re.M是必要的,因为您有^
,并且希望它在一行的开头匹配,而不仅仅是在整个字符串的开头。
你可以同时使用这两个标志,它会起作用,但re.DOTALL在这里没有做任何有用的事情。
如果不使用行断言,则不需要re.M
:
[ t]*([d:a-f/]+)[ t]*[VALID]s+([d:a-f/]+)
演示
但这可能会使正则表达式速度变慢
有了行断言,更快,[VALID]
两侧的字符类更具体:
演示2
Python演示(txt
设置为示例文本(:
>>> reg=re.compile(r'^[ t]*([d:a-f/]+)[ t]*[VALID]s+([d:a-f/]+)', flags=re.M)
>>> reg.findall(txt)
[('2001:11:2233::a1/24', '2001:cc11:22bb:0:2ec2:60ff:fe4f:feb2/64')]
使用YOUR regex或THIS regex,您不需要re.S
,它是使.
与n
匹配的标志。事实上,如果添加re.S
,您的正则表达式将只匹配第一个匹配项,因为.*
现在将匹配到字符串的末尾:
演示3