如何在python中提取URL的特定部分？

Question

我希望仅提取模式化URL的一部分：

https://<some_server>/server/rest/services/<that part I need>/<other data I do not need>/...

我只需要提取'rest / services /'之后的部分，第一部分可以更改，最后一部分可以更改，但是URL始终具有'rest / services /'，我需要紧随其后的是“ /...

Answer 1

您可以做的是通过拆分将该字符串转换为列表，然后访问第n个部分。

m = 'https://<some_server>/server/rest/services/<that part I need>/<other data I do not need>'
a = m.split('/')
a[6] //This is the part that you want

Answer 2

您可以尝试以下方法：

(?<=rest/services/) # look behind for this
[^/]+               # anything except a '/'

import re
rgxp = re.compile(r'(?<=rest/services/)[^/]+')
print re.findall(rgxp, text)