'function' object has no attribute 'as_view'


I am trying to use class based views, and get a strange error. The way I’m using the view seems to be the normal way:


from django.db import models
from django.utils import timezone

class Ingredient(models.Model):
    name        = models.CharField(max_length=255)
    description = models.TextField()

    def get_prices():
        purchases   = self.purchase_set.all()
        prices      = [purchase.price for purchase in purchases]


from django.shortcuts           import render, render_to_response, redirect
from django.http                import HttpResponse, HttpResponseRedirect
from django.views.generic.edit  import CreateView
from .models                    import Ingredient, Purchase

def IngredientCreateView(CreateView):
    model = Ingredient
    fields = ['all']


from django.conf.urls import patterns, include, url

from ingredients.views import IngredientCreateView

urlpatterns = patterns('',            
    url(r'^new_ingredient$',          IngredientCreateView.as_view(),             name='new-ingredient'),

I get

AttributeError at /ingredients/new_ingredient
'function' object has no attribute 'as_view'

I am on django 1.8.5. Why won’t this view work? Thank you

Asked By: codyc4321



IngredientCreateView should be a class.
So your views.py replace:

def IngredientCreateView(CreateView):


class IngredientCreateView(CreateView):
Answered By: Anush Devendra

IngredientCreateView is a function, not a class.

The following line

def IngredientCreateView(CreateView):

should be replace with

class IngredientCreateView(CreateView):
Answered By: falsetru

In my case, the problem was that I tried to use a @decorator on the class-based view as if it was a function-based view, instead of @decorating the class correctly.

EDIT: From the linked page, here is a way to apply @login_required to a class-based view:

from django.contrib.auth.decorators import login_required
from django.utils.decorators import method_decorator

@method_decorator(login_required, name='dispatch')
class ProtectedView(TemplateView):
Answered By: krubo

In addition to what is already said here, Check the file name and class name if it is same then you might have to import the class properly.

File name /api/A.py

class A:
//some methods

In your main class

//App main class
from api.A import A
Answered By: Shirish Singh

I faced the same problem but this solution worked for me..

in views.py file in viewclass you can use viewsets instead of CreateView

            from rest_framework import viewsets
            class YourClassView(viewsets.ModelViewSet):

in urls.py file you can use this routing pattern

          from django.conf.urls import url
          from rest_framework import routers

          router = routers.DefaultRouter()

          urlpatterns = [
               path('', include(router.urls)),
               path('admin/', admin.site.urls)
Answered By: Sureja_Hit
def Some_def_View(CreateView):

#should be replaced with

class SomeClassView(CreateView)
Answered By: Mahmoud Emad