15👍
>>> s="aaa?aaa"
>>> import re
>>> re.findall(r'aaa\?aaa', s)
['aaa?aaa']
The reason /aaa?aaa won’t match inside your URL is because a ? begins a new GET query.
So, the matchable part of the URL is only up to the first ‘aaa’. The remaining ‘?aaa’ is a new query string separated by the ‘?’ mark, containing a variable “aaa” being passed as a GET parameter.
What you can do here is encode the variable before it makes its way into the URL. The encoded form of ? is %3F.
You should also not match a GET query such as /?code=authenticationcode using regex at all. Instead, match your URL up to / using r'^$'. Django will pass the variable code as a GET parameter to the request object, which you can obtain in your view using request.GET.get('code').
2👍
You are not allowed to use ? in a URL as a variable value. The ? indicates that there are variables coming in.
Like: http://www.example.com?variable=1&another_variable=2
Replace it or escape it. Here’s some nice documentation.
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1👍
Django’s urls.py does not parse query strings, so there is no way to get this information at the urls.py file.
Instead, parse it in your view:
def foo(request):
code = request.GET.get('code')
if code:
# do stuff
else:
# No code!
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0👍
“How to match the ‘?’, is it special?”
Yes, but you are properly escaping it by using the backslash. I do not see where you have accounted for the leading forward slash, though. That bit just needs to be added in:
r'^/\?code=(?P<code>.*)$'
0👍
supress the regex metacharacters with []
>>> s
'/?code=authenticationcode'
>>> r=re.compile(r'^/[?]code=(.+)')
>>> m=r.match(s)
>>> m.groups()
('authenticationcode',)