我正在尝试使用基于类的视图制作索引页，该类视图将具有菜单栏，并通过该菜单栏将重定向到新页面，并将表单作为发布请求。但是无论我做什么，我总是会遇到相同的错误。
这是错误;
Exception in thread django-main-thread:
Traceback (most recent call last):
File "C:\Users\berat.berkol\anaconda3\lib\threading.py", line 926, in _bootstrap_inner
self.run()
File "C:\Users\berat.berkol\anaconda3\lib\threading.py", line 870, in run
self._target(*self._args, **self._kwargs)
File "C:\Users\berat.berkol\anaconda3\lib\site-packages\django\utils\autoreload.py", line 53, in wrapper
fn(*args, **kwargs)
File "C:\Users\berat.berkol\anaconda3\lib\site-packages\django\core\management\commands\runserver.py", line 118, in inner_run
self.check(display_num_errors=True)
File "C:\Users\berat.berkol\anaconda3\lib\site-packages\django\core\management\base.py", line 396, in check
databases=databases,
File "C:\Users\berat.berkol\anaconda3\lib\site-packages\django\core\checks\registry.py", line 70, in run_checks
new_errors = check(app_configs=app_configs, databases=databases)
File "C:\Users\berat.berkol\anaconda3\lib\site-packages\django\core\checks\urls.py", line 13, in check_url_config
return check_resolver(resolver)
File "C:\Users\berat.berkol\anaconda3\lib\site-packages\django\core\checks\urls.py", line 23, in check_resolver
return check_method()
File "C:\Users\berat.berkol\anaconda3\lib\site-packages\django\urls\resolvers.py", line 408, in check
for pattern in self.url_patterns:
File "C:\Users\berat.berkol\anaconda3\lib\site-packages\django\utils\functional.py", line 48, in __get__
res = instance.__dict__[self.name] = self.func(instance)
File "C:\Users\berat.berkol\anaconda3\lib\site-packages\django\urls\resolvers.py", line 589, in url_patterns
patterns = getattr(self.urlconf_module, "urlpatterns", self.urlconf_module)
File "C:\Users\berat.berkol\anaconda3\lib\site-packages\django\utils\functional.py", line 48, in __get__
res = instance.__dict__[self.name] = self.func(instance)
File "C:\Users\berat.berkol\anaconda3\lib\site-packages\django\urls\resolvers.py", line 582, in urlconf_module
return import_module(self.urlconf_name)
File "C:\Users\berat.berkol\anaconda3\lib\importlib\__init__.py", line 127, in import_module
return _bootstrap._gcd_import(name[level:], package, level)
File "<frozen importlib._bootstrap>", line 1006, in _gcd_import
File "<frozen importlib._bootstrap>", line 983, in _find_and_load
File "<frozen importlib._bootstrap>", line 967, in _find_and_load_unlocked
File "<frozen importlib._bootstrap>", line 677, in _load_unlocked
File "<frozen importlib._bootstrap_external>", line 728, in exec_module
File "<frozen importlib._bootstrap>", line 219, in _call_with_frames_removed
File "C:\Users\berat.berkol\SystemConsole\SystemConsole\urls.py", line 25, in <module>
path('', UserUpdateView.as_view(), name='home'),
AttributeError: 'function' object has no attribute 'as_view'
还有我项目的urls.py;
from django.contrib import admin
from django.urls import path
from django.conf.urls import include
from userupdate.views import UserUpdateView
app_name='userupdate'
urlpatterns = [
path('admin/', admin.site.urls),
path('login/',include('django.contrib.auth.urls')),
path('', UserUpdateView.as_view(), name='home'),
]
和我的views.py;
from django.shortcuts import render
from datetime import timedelta
from django.utils import timezone
from django.views.generic import TemplateView
from django.views.decorators.csrf import csrf_protect
from django.utils.decorators import method_decorator
from django.urls import reverse
import requests
from userupdate.forms import UserPassForm
from userupdate.forms import LoginForm
from django.views.generic import View
from django.contrib.auth.views import LoginView
from django.contrib.auth.mixins import LoginRequiredMixin
@csrf_protect
class UserUpdateView(LoginRequiredMixin,View):
emplate_name='home.html'
def get(self, request, *args, **kwargs):
return render(request,'home.html')
#@method_decorator(login_required)
def dispatch(self, request, *args, **kwargs):
return super(UserUpdateView, self).dispatch(self, request, *args, **kwargs)
@csrf_protect
class LoginUser(LoginView):
template_name = 'home.html' # your template
from_class = LoginForm() # your form
def get_success_url(self):
'''Here the part where you can implement your login logic'''
now = timezone.now()
# Get current day date object
# like: 12/02/2019 00:00:00
today = now.replace(minute=0).replace(second=0).replace(microsecond=0)
# Get the client from the user object
client = self.request.user.cli
# Get all the user today's logins and count them
client_logins = models.ClientLogins.objects.filter(
client=client,
date__gte=today,
date__lte=today + timedelta(days=1)
).count()
if client_logins < 1: # Or: if not client_logins:
# create a login tracker record
models.ClientLogins.objects.create(
client=client,
date=now # Store the date where the user logged in the website
)
return reverse_lazy('home')
# Or redirect to: settings.LOGIN_REDIRECT_URL
request.session.get_expire_at_browser_close()
return super().get_success_url()
@csrf_protect
class passResetView(View):
def get(self,request):
form=UserUpdateForm()
return render('basarili')
def post(self,request):
passResetuser=self.model.objects.get(pk=2)
form=UserUpdateForm(request.POST,instance=passResetuser)
if form.is_valid():
username=form.cleaned_data('username')
password=form.cleaned_data('password')
number=form.cleaned_data('number')
message=form.cleaned_data('message')
context={'form':form}
return render(request,'home.html',context)
我不知道是否需要，但这是我的应用程序的urls.py;
from django.urls import path, re_path
from django.contrib.auth.views import LoginView
from userupdate.views import HomeView
from django.contrib.auth import views as auth_views
app_name='userupdate'
urlpatterns = [
path('', views.UserUpdateView.as_view(template_name='home.html'), name='home'),
path('login/',
LoginView.as_view(
template_name='login.html'),
name="login"),
#path('passResetView',views.passResetView.as_view(template_name="home.html"),name='passResetView'),
path('passreset/',
passResetView.as_view(
template_name='passreset.html'),
name="PasswordReset"),
]
我尚未配置表单页面，因此暂时将其忽略。
我将使用FormView构建它们。
通过我也尝试了TemplateView的方式，但是我没有帮助它产生相同的错误。
答案 0 :(得分：2)
您不能在基于类的视图上使用@csrf_protect，或者至少不能直接使用。该装饰器将返回一个函数，并且该函数当然没有.as_view()方法。
您可以使用@method_decorator [Django-doc]，它将装饰类：
from django.utils.decorators import method_decorator
from django.views.decorators.csrf import csrf_protect
@method_decorator(csrf_exempt, name='dispatch')
class UserUpdateView(LoginRequiredMixin,View):
template_name = 'home.html'
# …话虽这么说，豁免CSRF通常不是一个好主意，因为这会使您的视图容易受Cross-Site Request Forgery [wiki]的攻击。</ p>