正则表达式:只匹配一次

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问题描述

我有一个包含多个 ip 地址的字符串以及一些随机的东西.比如这个:

21/Jun/2018:01:15:38 +0000 188.79.169.152 157.52.69.50 443 - - GET/157.52.69.30 157.52.69.10

得到这个正则表达式:

[0-9]+(?:\.[0-9]+){3}

问题是这匹配多次，但我不需要那个.这个正则表达式缺少什么，所以它只会匹配一次?

谢谢，

解决方案

您可以使用 re.match 从字符串的开头匹配 ptrn因此，只需在模式的开头添加一个 .*，我们就可以匹配从字符串开头到第一个 IP 地址的所有内容

>>>进口重新>>>s = "21/Jun/2018:01:15:38 +0000 188.79.169.152 157.52.69.50 443 - - GET/157.52.69.30 157.52.69.10">>>>>>ptrn = r'.*?([0-9]+(?:\.[0-9]+){3})'>>>re.match(ptrn, s).groups()[0]'188.79.169.152'

I've got a string with multiple ip addresses together with some random stuff. For example like this one:

21/Jun/2018:01:15:38 +0000    188.79.169.152    157.52.69.50    443    -    -    GET / 157.52.69.30 157.52.69.10

And got this regex:

[0-9]+(?:\.[0-9]+){3}

Problem is that this matches multiple times, but I don't need that. What is missing with this regex so it will only match once?

Thanks,

解决方案

You can use re.match which matches a ptrn from the beginning of a stringSo just by adding a .* to the start of your pattern, we can match everything from the beginning of the string to the first ip address

>>> import re
>>> s = "21/Jun/2018:01:15:38 +0000    188.79.169.152    157.52.69.50    443    -    -    GET / 157.52.69.30 157.52.69.10"
>>>
>>> ptrn = r'.*?([0-9]+(?:\.[0-9]+){3})'
>>> re.match(ptrn, s).groups()[0]
'188.79.169.152'

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