How to return 404 page intentionally in django

Question

I made custom 404 page in django. And I'm trying to get 404 error page intentionally.

myproject/urls.py:

from website.views import customhandler404, customhandler500, test

urlpatterns = [
    re_path(r'^admin/', admin.site.urls),
    re_path(r'^test/$', test, name='test'),
]
handler404 = customhandler404
handler500 = customhandler500

website/views.py

def customhandler404(request):
    response = render(request, '404.html',)
    response.status_code = 404
    return response


def customhandler500(request):
    response = render(request, '500.html',)
    response.status_code = 500
    return response

def test(request):
    raise Http404('hello')

But when I go 127.0.0.1:8000/test/ , It seems to return 500.html

And terminal says:

[24/Mar/2018 22:32:17] "GET /test/ HTTP/1.1" 500 128

How can I intentionally get 404 page?

Answer 1

When you set debug to False, you don't have a custom handler, and the status code of the response is 404, the 404.html (if present) in your base template directory is used. To return a response with a 404 status, you can simply return an instance of django.http.HttpResponseNotFound. The reason you got a 500 is because you raised an error instead of returning a response. So, your test function can be simply modified to this

from django.http import HttpResponseNotFound
def test(request):
    return HttpResponseNotFound("hello")         

Update:

So it turned out that the reason you are getting a 500 error was not that you raised an exception, but having incorrect function signatures. When I answered this question more than half a year ago I forgot that django catches HTTP404 exception for you. However, the handler view has different signatures than the normal views. The default handler for 404 is defaults.page_not_found(request, exception, template_name='404.html'), which takes 3 arguments. So your custom handler should actually be

def customhandler404(request, exception, template_name='404.html'):
    response = render(request, template_name)
    response.status_code = 404
    return response

Although, in this case, you may as well just use the default handler.