我还是Django的新手，似乎无法解决此错误（我在这里读过类似的文章没有运气）。 urls.py中的代码粘贴如下：
from django.conf.urls import include, url
from django.contrib import admin
# Add this import
from django.contrib.auth import views
from log.forms import LoginForm
urlpatterns = [
url(r'^admin/', admin.site.urls),
url(r'^', include(('dashboard.urls', "dashboard"), namespace='dashboard')),
url(r'^login/$', views.LoginView, {'template_name': 'login.html', 'authentication_form': LoginForm}, name='login'),
url(r'^logout/$', views.LogoutView, {'next_page': '/'}),
]
完整跟踪如下：
Internal Server Error: /login/
Traceback (most recent call last):
File "C:\Python36-32\lib\site-packages\django\core\handlers\exception.py", line 34, in inner
response = get_response(request)
File "C:\Python36-32\lib\site-packages\django\core\handlers\base.py", line 127, in _get_response
response = self.process_exception_by_middleware(e, request)
File "C:\Python36-32\lib\site-packages\django\core\handlers\base.py", line 125, in _get_response
response = wrapped_callback(request, *callback_args, **callback_kwargs)
TypeError: __init__() takes 1 positional argument but 2 were given
[21/Aug/2018 00:20:27] "GET /login/?next=/ HTTP/1.1" 500 64213
关于如何解决此问题的任何想法？还是什么可能导致此错误？
答案 0 :(得分：1)
简而言之：您需要使用 .as_view() 将基于类的视图“转换”为可以在{{ 1}}。
LoginView [Django-doc]和LogoutView [Django-doc]是基于类的视图，为了使其在urls.py中可调用，您需要使用{{1} }：
urls.py基于类的视图包含充当包装器的功能，并且每次初始化时都如此。如果直接使用as_view，则将调用from django.conf.urls import include, url
from django.contrib import admin
# Add this import
from django.contrib.auth import views
from log.forms import LoginForm
urlpatterns = [
url(r'^admin/', admin.site.urls),
url(r'^', include(('dashboard.urls', "dashboard"), namespace='dashboard')),
url(r'^login/$', views.LoginView.as_view(), {'template_name': 'login.html', 'authentication_form': LoginForm}, name='login'),
url(r'^logout/$', views.LogoutView.as_view(), {'next_page': '/'}),
]类的构造函数。尽管有一些额外的逻辑，但有可能返回一个LoginView，但它不是很优雅（构造LoginView时您不会期望HttpResponse），而且还会返回使视图的子类化（这是此类基于类的视图可以节省大量工作的原因之一）非常麻烦。