How to create a urlpattern with an arbitrary number of parameters of the same type?

Question

Django

How to compose a url that allows you to pass several parameters to the view, the number of which is not known in advance?
It is clear that the classical approach is as follows:

from django.conf.urls import url

from .views import desktop
from .views import users

urlpatterns = [
    url(r'^desktop/$', my_view),
    url(r'^desktop/users/$, users)
]

I want to make a class-based-view that will render the necessary templates, which are beautifully arranged in folders inside the templates directory. However, each folder can have subfolders. The view will analyze the number of received parameters ( *args ) and, depending on this, determine the nesting depth and the path to the desired html file.

Answer 1

Comrades from StackOverflow recommend this solution to the issue:

urlpatterns = [
    url(r'^(?P<path>[-\w/]+)/$', TemplateView.as_view()),
]

Answer 2

not a solution but look seriously towards regex