Passing user auth status in view

Question

I am using Django Cookiecutter as template. And by default in urls.py there is following url:

url(r'^about/$', TemplateView.as_view(template_name='pages/about.html'), name='about'),

and because i need to pass some additional parameters in it i wanted to use it as:

url(r'^about/$', index, name='about'),

Where index is from my views, but when i use the view, my system does not recognise that user is logged in, should i somehow pass user in my index too or what am i doing wrong?

What i am doing in my view:

def index(request):
    return render_to_response('pages/about.html', {
        'categories': Category.objects.all(),
        'posts': Post.objects.all()[:5]
    })

please elaborate how are you checking the user in the template? — Exprator, Jul 18 '17 at 10:19
I guess you have a decorator in the view requiring authentication or something? User data is normally added the to `request` object in middleware. — Grimmy, Jul 18 '17 at 10:19
All is from boilerplate about authentication. So i don't really require that user is logged in, but when i use view function user object is not passed like it is in TemplateView. — HyperX, Jul 18 '17 at 10:21
https://docs.djangoproject.com/en/1.11/ref/templates/api/#django-contrib-auth-context-processors-auth — bruno desthuilliers, Jul 18 '17 at 10:23
Context processor for auth is there so this is not a problem. — HyperX, Jul 18 '17 at 10:26

score 1 · Accepted Answer · answered Jul 18 '17 at 10:31

I solved it by using render instead of render to response so the method after change looks like this:

def index(request):
    categories = Category.objects.all()
    posts = Post.objects.all()[:5]
    context = {'categories': categories, 'posts': posts}
    return render(request, 'pages/home.html', context)

Passing user auth status in view

1 Answers1