How to remove query string from a url?

You can split on something that doesn't exist in the string, you'll just get a list of one element, so depending on your goal, you could do something like this to simplify your existing code:

url = url.split('?')[0].split('#')[0]

Not saying this is the best way (furl is a great solution), but it is a way.

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'

The very helpful library furl makes it trivial to remove both query and fragment parts:

>>> furl.furl("https://hi.com/?abc=def#ghi").remove(args=True, fragment=True).url
https://hi.com/