as_view() 采用 1 个位置参数，但给出了 2 个

Question

帮我解决这个问题url.pyurlpatterns = [    url(r'login/$',auth_views.LoginView.as_view(template_name="accounts/login.html"),name='login'),    url(r'logout/$',auth_views.LogoutView.as_view(),name='logout'),    url(r'signup/$',views.SignUp.as_view(),name='signup'),]views.pyclass SignUp(CreateView):    form_class = forms.UserCreateForm    success_url = reverse_lazy("login")    template_name = 'accounts/signup.html'这是我得到的错误回溯Traceback: File "C:\Users\DELL\anaconda3\envs\MyDjangoEnv\lib\site-packages\django\core\handlers\exception.py",line 34, in inner response = get_response(request) File "C:\Users\DELL\anaconda3\envs\MyDjangoEnv\lib\site-packages\django\core\handlers\base.py",line 115, in _get_response response = self.process_exception_by_middleware(e, request)File "C:\Users\DELL\anaconda3\envs\MyDjangoEnv\lib\site-packages\django\core\handlers\base.py",line 113, in _get_response response = wrapped_callback(request, *callback_args, **callback_kwargs) –请帮忙

Answer 1

我不确定，但在 urls.py 中试试这个模式是怎样的 path('signup/',views.SignUp.as_view( success_url=reverse_lazy('login'), name='signup' ),name='signup') from django.urls import reverse_lazy