How to remove query string from a url?

Question

I have the following URL:

https://stackoverflow.com/questions/7990301?aaa=aaa
https://stackoverflow.com/questions/7990300?fr=aladdin
https://stackoverflow.com/questions/22375#6
https://stackoverflow.com/questions/22375?
https://stackoverflow.com/questions/22375#3_1

I need URLs for example:

https://stackoverflow.com/questions/7990301
https://stackoverflow.com/questions/7990300
https://stackoverflow.com/questions/22375
https://stackoverflow.com/questions/22375
https://stackoverflow.com/questions/22375

My attempt:

url='https://stackoverflow.com/questions/7990301?aaa=aaa'
if '?' in url:
    url=url.split('?')[0]
if '#' in url:
    url = url.split('#')[0]

I think this is a stupid way

Answer 1

In your example you're also removing the fragment (the thing after a #), not just the query.

You can remove both by using urllib.parse.urlsplit, then calling ._replace on the namedtuple it returns and converting back to a string URL with urllib.parse.unsplit:

from urllib.parse import urlsplit, urlunsplit

def remove_query_params_and_fragment(url):
    return urlunsplit(urlsplit(url)._replace(query="", fragment=""))

Output:

>>> remove_query_params_and_fragment("https://stackoverflow.com/questions/7990301?aaa=aaa")
'https://stackoverflow.com/questions/7990301'
>>> remove_query_params_and_fragment("https://stackoverflow.com/questions/7990300?fr=aladdin")
'https://stackoverflow.com/questions/7990300'
>>> remove_query_params_and_fragment("https://stackoverflow.com/questions/22375#6")
'https://stackoverflow.com/questions/22375'
>>> remove_query_params_and_fragment("https://stackoverflow.com/questions/22375?")
'https://stackoverflow.com/questions/22375'
>>> remove_query_params_and_fragment("https://stackoverflow.com/questions/22375#3_1")
'https://stackoverflow.com/questions/22375'