Regex for LinkedIn profile

I have want to have an function that replace a LinkedIn profile that i found in a string.

example:

• You can find my linkedin https:www.linkedin.com/in/kim-zand-3126573/

After using the function it should be:

• You can find my linkedin [Linkedin]

strA ='[Linkedin]'

Def linkedin(sentence):
urlReg = "^https?:\/\/?(w{3}.)? linkedin\.\/.$"
res = re.search(urlReg, sentence)
print(res)
if res != None:
    ## replace urls
    sentence= re.sub(urlReg,strA, sentence)
return (sentence)

When i print (res) it does not taking the string in, i think my Regex is not correct

>Solution :

A few notes about the pattern that you tried:

• There is no // in https:www.linkedin.com/in/kim-zand-3126573/ (maybe a typo?)
• linkedin\.\/ does not match linkedin.com as \/ matches /
• There is no space in the example string, that is expected in the regex before linkedin\.
• The text is not at the start of the string ^ while there is an anchor
• You have to escape the dot to match it literally
• The second forward slash in // does not have to be optional and you don’t have to escape the forward slash

You could for example match:

\bhttps?://www\.linkedin\.com/\S*

Or if there is really no double forward slash in the data:

\bhttps?:www\.linkedin\.com/\S*

And then replace with [Linkedin]

See a regex demo

Or a bit more specific match for the given example, matching the -digits part before the last /

\bhttps?://www\.linkedin\.com\/[^\s/]+/[^/\s]+-\d+/(?!\S)

See another regex demo

Example:

import re

strA = '[Linkedin]'
urlReg = re.compile(r"\bhttps?://www\.linkedin\.com/\S*")

def linkedin(sentence):
    return urlReg.sub(strA, sentence)

result = linkedin(r"You can find my linkedin https://www.linkedin.com/in/kim-zand-3126573/")
print(result)

Output

You can find my linkedin [Linkedin]