Reputation: 946
I am following the Django tutorial. I have a /films
page and /films/id
page displaying successfully. I would like to divert slightly from the tutorial and one step of this is to display the content which is currently appearing in /films
in the website home e.g. http://127.0.0.1:8000/
I currently have:
index.py:
def index(request):
latest_film_list = Film.objects.order_by('-pub_date')[:5]
context = {'latest_film_list': latest_film_list}
return render(request, 'films/index.html', context)
index.html:
{% if latest_film_list %}
<ul>
{% for film in latest_film_list %}
<li><a href="/films/{{ film.id }}/">{{ film.title }}</a></li>
{% endfor %}
</ul>
{% else %}
<p>No films are available.</p>
{% endif %}
films/urls.py:
urlpatterns = [
path('', views.index, name='index'),
# # ex: /films/5/
path('<int:film_id>/', views.detail, name='detail'),
# # ex: /films/5/results/
path('<int:film_id>/results/', views.results, name='results'),
# # ex: /films/5/vote/
path('<int:film_id>/vote/', views.vote, name='vote'),
]
mysite.urls.py
urlpatterns = [
url(r'^films/', include('films.urls')),
url(r'^admin/', admin.site.urls),
]
I realise I probably just need to add/adjust the films/urls.py file but have been unsuccessful so far. Home currently displays a 404.
Upvotes: 0
Views: 39
Reputation: 345
You can simply change your url as follows:
From this:
url(r'^films/', include('films.urls'))
To this:
url(r'^', include('films.urls'))
Upvotes: 1