django-templates - LogoutView 中的模板名称不适用于 django2.1 python3.7

Question

下面是 django.contrib.auth.views.LogoutViews 中的默认设置，

template_name = 'registration/logged_out.html'

我像这样配置我的应用程序的 urls.py:

from django.urls import path
from . import views

from django.conf import settings

from django.contrib.auth.views import LoginView, LogoutView

app_name = 'account'
urlpatterns = [
    #path("login/", views.user_login, name="user_login"),
    path("login/", LoginView.as_view(), name="user_login"),
    path("nlogin/", LoginView.as_view(), {"template_name":"account/login.html"}),
    path("logout/", LogoutView.as_view(), name="user_logout"),
    path("logoutt/", LogoutView.as_view(), {"template_name":"account/logout.html"}),
]

"template_name":"account/login.html"工作正常，但 "template_name":"account/logout.html"似乎没有区别，我的代码有什么问题？

Answer 1

当您使用基于类的变体时，您通过 .as_view 将设置传递给 View 。 (所谓的 **initkwargs )方法:

from django.urls import path
from . import views

from django.conf import settings

from django.contrib.auth.views import LoginView, LogoutView

app_name = 'account'

urlpatterns = [
    #path("login/", views.user_login, name="user_login"),
    path("login/", LoginView.as_view(), name="user_login"),
    path("nlogin/", LoginView.as_view(template_name='account/login.html')),
    path("logout/", LogoutView.as_view(), name="user_logout"),
    path("logoutt/", LogoutView.as_view(template_name='account/logout.html')),
]

否则参数将在 self.kwargs 中结束，并且基于类的 View 不会检查这些。

documentation on the LoginView [Django-doc]提到了这一点以及您可以作为 **initkwargs 传递的参数列表.