如何在python上使用正则表达式获取链接的特定部分

Question

I'm inputting a lot of links from linked-in profiles into a program that will get me the id of those linked-in profiles. (Links are the strings, clicking on most of them will take you nowhere)

Example 1: " https://www.linkedin.com/in/facundo-b-barber%C3%A1-86bb41187/ "

Example 2: " https://www.linkedin.com/in/facundo-b-barber%C3%A1-86bb41187/sometext "

If I input either of those examples the result will be: "facundo-b-barber%C3%A1-86bb41187" The problem I run into is when I have something like this:

Example 3: " https://www.linkedin.com/in/facundo-b-barber%C3%A1-86bb41187/sometext/anothertext/ "

Where the output is: "facundo-b-barber%C3%A1-86bb41187/sometext"

I've tried using re module in this function:

def get_in(url):
    parsed = parse.urlparse(url)
    lin = parsed.path
    lin = re.search(r'/in/(.*)/', lin).group(1)
    print(lin)
    return lin

I want to get the id only and remove everything else in front and behind.

Answer 1

This should work ->

url.split('/')[4]

Examples:

>>> url =  "https://www.linkedin.com/in/facundo-b-barber%C3%A1-86bb41187/sometext/anothertext/"
>>> url.split('/')[4]
'facundo-b-barber%C3%A1-86bb41187'

>>> url = "https://www.linkedin.com/in/facundo-b-barber%C3%A1-86bb41187/sometext"
>>> url.split('/')[4]
'facundo-b-barber%C3%A1-86bb41187'

>>> url = "https://www.linkedin.com/in/facundo-b-barber%C3%A1-86bb41187/"
>>> url.split('/')[4]
'facundo-b-barber%C3%A1-86bb41187'