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Python Django,'Django.contrib.auth.views'没有“login”属性_Python_Django_Django Views_Django Authentication - Fatal编程技术网

Python Django,'Django.contrib.auth.views'没有“login”属性

Python Django,'Django.contrib.auth.views'没有“login”属性,python,django,django-views,django-authentication,Python,Django,Django Views,Django Authentication,当我试图从django.contrib.auth导入视图时,我遇到了这个属性错误,我想使用它们的身份验证方法 from django.contrib import admin from django.urls import path from django.conf.urls import url,include from django.contrib.auth import views urlpatterns = [ url(r'', include('blog.urls')),

当我试图从django.contrib.auth导入视图时,我遇到了这个属性错误,我想使用它们的身份验证方法

from django.contrib import admin
from django.urls import path
from django.conf.urls import url,include
from django.contrib.auth import views

urlpatterns = [
    url(r'', include('blog.urls')),
    url(r'^accounts/login/$', views.login, name='login'),
    url(r'^accounts/logout/$', views.logout, name='logout', kwargs= 
{'next_page':'/' }),
    path('admin/', admin.site.urls),
]
AttributeError:模块“django.contrib.auth.views”没有“login”属性

urlr“^accounts/login/$”,views.login,name='login',

auth.view模块没有登录功能

试试这个Django 2.2

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


url(r'^accounts/login/$', LoginView.as_view(), name='login'),
url(r'^accounts/logout/$', LogoutView.as_view(), name='logout'),

or 

url(r'^accounts/login/$', logout_then_login, name='login'),