Question

帮我解决这个问题

url.py

urlpatterns = [
    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.py

class SignUp(CreateView):
    form_class = forms.UserCreateForm
    success_url = reverse_lazy("login")
    template_name = 'accounts/signup.html'

这是我得到的错误跟踪

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) –

请帮助

Answer 1

我不确定，但可以在urls.py中尝试一些模式 path('signup/',views.SignUp.as_view( success_url=reverse_lazy('login'), name='signup' ),name='signup') from django.urls import reverse_lazy

as_view（）接受1个位置参数，但给出了2个

1 个答案: