Devolver el resultado de una función como vista:

# En urls:
urlpatterns = [
    path('videos/', views.VideoListView.as_view(), name='videos'),
]
 
# En views:
from django.http import HttpResponse
def index(request):
    return HttpResponse('Hey yo!')

If you don't want to pass a pk or slug in your url, you'll need to override the get_object method and get your object another way.

path('profile/<username>', views.UserDetailView.as_view(), name='profile')

class UserDetailView(generic.DetailView):
    model = apps.get_model('openawards', 'User')
    username = ''

    def dispatch(self, request, *args, **kwargs):
        super().dispatch(request, *args, **kwargs)
        self.username = kwargs.get('username', '')

    def get_object(self):
        return get_object_or_404(get_user_model(), username=self.username)

Me pregunto si hay otra forma mejor de sacar el parámetro de la url en vez de sobreescribiendo el dispatch. Si!! Siempre puedes acceder a los kwargs de la request:

return get_object_or_404(get_user_model(), username=self.kwargs.get('username', ''))

Si los datos viniesen por query string:

username = self.request.GET.get('username', '')

En el urls.py:

login_required(ViewSpaceIndex.as_view(..))

Usando LoginRequiredMixin:

from django.contrib.auth.mixins import LoginRequiredMixin
class MyView(LoginRequiredMixin, View):
    login_url = '/login/'
    redirect_field_name = 'redirect_to'

Con el method_decorator:

from django.utils.decorators import method_decorator
@method_decorator(login_required, name='dispatch')
class ViewSpaceIndex(TemplateView):
    template_name = 'secret.html'