CODEWITHSUNDEEP
pythonpython-3.xlistinputfindall
541daw35d
541daw35d

Reputation: 151

Python findall doesn't return results even on a supposed proper match

I am trying to filter out just links from an input that the user provides. However currently I am having trouble with Python's findall as it does not return any results.

There might be a mistake in the regular expression itself since that's not a part of my actual code and I copied that in another post but I tried to check it with my current understanding of regular expressions and didn't see an obvious issue.

These are the example links that I am trying to filter :

https://ncs.io/symbolism, https://soundcloud.com/nocopyrightsounds/electro-light-symbolism-ncs-release, https://www.youtube.com/watch?v=__CRWE-L45k, https://www.deezer.com/track/93333734?app_id=140685, https://open.spotify.com/album/1BxVIZuG2N1ReBB0T5nVTr?highlight=spotify:track:2zVJlAEB0ublkqJMIn43AE

To get the results I am using a simple function which should directly return the result :

def get_urls():
    # Gets urls from the provided string
    urls = input("Stream urls: ")
    urls = findall(r'(https?://[^\s]+)', urls)
    return urls

Which I then properly call :

if __name__ == '__main__':
    print(get_urls())

I am importing just findall from re since that's the only part I use :

from re import findall

There isn't really much that can break in this small code and I myself am extremely confused as to where I screwed up. If you see me mistake please help me out! Thanks for any help with this!

Upvotes: 0

Views: 132

Answers (1)

Marcel Preda
Marcel Preda

Reputation: 1205

It works for me

from re import findall

def get_urls():
    # Gets urls from the provided string
    urls = input("Stream urls: ")
    urls = findall(r'(https?://[^\s,]+)', urls)
    return urls

if __name__ == '__main__':
    print(get_urls())

Upvotes: 1

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