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IP 正则表达式,所有站点都没有子网

[英]IP Regex without subnet for all sites

As a part of a project of my company, I need to extract IP addresses that don't include subnetting (eg 196.82.1.12/24) from some websites.作为我公司项目的一部分,我需要从某些网站中提取不包括子网划分的 IP 地址(例如 196.82.1.12/24)。

If the address contains subnetting, I don't want to grab the part proceeding the subnetting but not taking it at all.如果地址包含子网划分,我不想抓住继续进行子网划分但根本不采取它的部分。

for example on the following case:例如在以下情况下:

<td>212.179.35.154</td>
<td>200.139.97.126/24</td>
<td>"201.139.97.126"</td>
<td>F5 BIG-IP</td>
<td>unknown</td>
<td class="date">26-Feb-2011</td>

The desired output would be:所需的 output 将是:

212.179.35.154 212.179.35.154

201.139.97.126 201.139.97.126

Please note that some lines include quotes surrounding the IP address however since there is no following /NUMBER they are valid.请注意,有些行包含围绕 IP 地址的引号,但是由于没有以下 /NUMBER,因此它们是有效的。

I'm trying to find an appropriate regex for days now such as:我现在正在尝试找到合适的正则表达式,例如:

(<td>(\d+\.){3}\d+<\/td>)
\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3}[^\/]

However, all seem to have a flaw within them.然而,所有这些似乎都存在缺陷。

Thanks in advance!提前致谢!

For me it looks like task where negative lookahead will be useful.对我来说,它看起来像是负前瞻很有用的任务。 I would do:我会做:

import re
txt = '''<td>212.179.35.154</td>
<td>200.139.97.126/24</td>
<td>"201.139.97.126"</td>
<td>F5 BIG-IP</td>
<td>unknown</td>
<td class="date">26-Feb-2011</td>'''
pattern = r"\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3}(?![0-9/])"
found = re.findall(pattern, txt)
print(found)

Output: Output:

['212.179.35.154', '201.139.97.126']

By using negative lookahead (?![0-9/]) we say: exclude matches if they are followed by 0 or 1 or 2 or 3 or 4 or 5 or 6 or 7 or 8 or 9 or / .通过使用负前瞻(?![0-9/])我们说:如果匹配项后面跟着0123456789/ ,则排除匹配项。 Note that including digits is crucial here, because if you specify only / one of matches would be:请注意,此处包含数字至关重要,因为如果您仅指定 / 匹配项之一将是:

200.139.97.12

(note missing 6 at end) (注意最后少了6

You can use a negative lookahead assertion , by using the pattern syntax (?....) , like this:您可以通过使用模式语法(?....)来使用否定的前瞻断言,如下所示:

import re

s = """
<td>212.179.35.154</td>
<td>200.139.97.126/24</td>
<td>"201.139.97.126"</td>
<td>F5 BIG-IP</td>
<td>unknown</td>
<td class="date">26-Feb-2011</td>
"""

pattern = r"\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3}(?!\d*\/)"

print(re.findall(pattern,s))

Output: Output:

['212.179.35.154', '201.139.97.126']

The (?!\d*\/) part tells it "don't match the previous pattern if it is followed by any digits and a forward slash". (?!\d*\/)部分告诉它“如果后面跟着任何数字和正斜杠,则不匹配前一个模式”。
(the \d* part is because otherwise it will match 200.139.97.12 (without the 6 ) out of 200.139.97.126/24 ) \d*部分是因为否则它将匹配200.139.97.12 200.139.97.126/24没有6

small note: your original pattern will match more than just legal IP addresses, but I went with your way.小提示:您的原始模式将不仅仅匹配合法的 IP 地址,但我选择了您的方式。

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