How to handle dynamic root slug?

Question

Django

Hello. I started learning Django and ran into such a problem, now I'm trying to make a simple product catalog and I wanted to do it beautifully, so that the urls were without the word catalog at the beginning. For example something like this:
Product categories:
/telefoni/knopochniye/
/telefoni/knopochniye/samsung
Product page:
/telefoni/knopochniye/beliy-telefon-androyd
Interested in how to process such URLs correctly, because in all the examples that I saw, there is some kind of "static" part that you can hook onto, for example:

urlpatterns = [
    path('catalog/brand/samsung', views.brand_page, name='brand_page'),
    ...
]

And if you go further, and for example, want to have text pages in the future, the URLs of which are stored in the database, then it becomes generally unclear how to do routing correctly.

Answer 1

Insert directory - writing your own routing is not for beginners

Answer 2

Try to start with three views:

from django.urls import include, re_path

urlpatterns = [
    re_path(r'^(?P<catalog>\w+)/$', views.catalog_view, name='catalog'),
    re_path(r'^(?P<catalog>\w+)/(?P<brand>\w+)/$', views.brand_view, name='brand'),
    re_path(r'^(?P<catalog>\w+)/(?P<brand>\w+)/(?P<product>\w+)/$', views.product_view, name='product'),
    ...
]

Views will accept slug as parameters, like so:

def product_view(request, catalog, brand, product):
    # ...