How to decompose URL into model and parameters

[9dev...@gmail.com]

I want to parse an application URL just like Django does.

url = 'http://www.example.com/path/to/myview/123'
view, params = decompose(url)
# now view="MyView", params=('123',)

How can it be done?

[Collin Anderson]

from django.core import urlresolvers

match = urlresolvers.resolve(urlsplit(url).path)
view = match.func
params = math.args

https://docs.djangoproject.com/en/dev/ref/urlresolvers/#resolve

[Collin Anderson]

It raises an Http404 if it doesn't exist. Is that what's going on?

[9dev...@gmail.com]

Yes, I figured it out. Thank you.