django，未定义名称“IndexView”

Question

我正在关注本教程。目前我处于这一点，但是当我启动服务器python manage.py runserver 0.0.0.0:8000并在浏览器中打开 url 时，我收到以下错误：

name 'IndexView' is not defined

这是我的 urls.py

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

from rest_framework_nested import routers
from authentication.views import AccountViewSet

router = routers.SimpleRouter()
router.register(r'accounts', AccountViewSet)

urlpatterns = patterns(
    '',
    url(r'^admin/', include(admin.site.urls)),

    url(r'^api/v1/', include(router.urls)),
    url('^.*$', IndexView.as_view(), name='index'),
)

我不知道如何解决这个问题，因为我从未见过自己甚至在IndexView某处声明过。如果你们能给我一些关于这个的建议，那就太棒了。

编辑：

我的意见.py

from django.shortcuts import render

# Create your views here.

from rest_framework import permissions, viewsets

from authentication.models import Account
from authentication.permissions import IsAccountOwner
from authentication.serializers import AccountSerializer

class AccountViewSet(viewsets.ModelViewSet):
    lookup_field = 'username'
    queryset = Account.objects.all()
    serializer_class = AccountSerializer

    def get_permissions(self):
        if self.request.method in permissions.SAFE_METHODS:
            return (permissions.AllowAny(),)

        if self.request.method == 'POST':
            return (permissions.AllowAny(),)

        return (permissions.IsAuthenticated(), IsAccountOwner(),)

    def create(self, request):
        serializer = self.serializer_class(data=request.data)

        if serializer.is_valid():
            Account.objects.create_user(**serializer.validated_data)

            return Response(serializer.validated_data, status=status.HTTP_201_CREATED)

        return Response({
            'status': 'Bad request',
            'message': 'Account could not be created with received data.'
        }, status = status.HTTP_400_BAD_REQUEST)

Answer 1

您必须创建它IndexView并将其导入到您的urls.py. 目前口译员抱怨，因为在urls.py IndexView未知。要创建新视图，您应该在 中创建一个新类views.py，例如：

from django.views.generic.base import TemplateView

class IndexView(TemplateView):
    template_name = 'index.html'

ps：请阅读Django官方文档，非常好！