Challenge 2: Wire It Into the Project — Possible Solution
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# mysite/urls.py — the project's root routing table

from django.contrib import admin
from django.urls import path, include

urlpatterns = [
    path("admin/", admin.site.urls),
    path("books/", include("catalog.urls")),
    path("accounts/", include("accounts.urls")),
]

FINAL FULL PATHS FOR EACH ROUTE FROM CHALLENGE 1
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Given catalog.urls is included under the "books/" prefix:

- path("", views.book_list, name="book-list")
    -> combined with the "books/" prefix -> full path: /books/

- path("<int:book_id>/", views.book_detail, name="book-detail")
    -> full path: /books/<book_id>/  (e.g. /books/42/)

- path("category/<str:slug>/", views.by_category, name="book-category")
    -> full path: /books/category/<slug>/  (e.g. /books/category/sci-fi/)

WHY THIS WORKS
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- include("catalog.urls") tells Django: "for anything under books/, hand
  the REST of the path off to catalog's own urlpatterns list, stripping
  the books/ prefix before matching against catalog's patterns." This is
  why catalog/urls.py itself never mentions "books/" anywhere — it's
  written as if it owned the root, and the project decides what prefix
  to mount it under.
- Because the prefix is decided entirely in the project's root urls.py,
  the SAME catalog app could be mounted under a different prefix (say
  "library/") in another project, or even twice under two different
  prefixes in the same project, without changing a single line inside
  catalog/urls.py.
- admin.site.urls is Django's own built-in admin routing (Chapter 8 covers
  the admin interface itself) — it's included the same way any app's URLs
  would be, just pointing at Django's internal admin app instead of one
  of your own.