django覆盖auth登录方法

Question

我想重写django.contrib.auth的__init__.py中的login函数

我做了以下步骤:

urls.py

url(r'^login/$', 'my_login'),

views.py

from django.contrib import auth

def login(request, user, backend=None):
   # do some stuff

settings.py

   INSTALLED_APPS = [
   #'django.contrib.auth',
    'my_auth_app'
   ]

但我觉得这是错误的做法。

实际上我想使用覆盖的login方法获得默认的contrib.auth

Answer 1

完成了这个方法。

url.py

# separate my custom login from django default auth
path('accounts/login/', CustomLoginView.as_view(), name='login'), 
path('accounts/', include('django.contrib.auth.urls')),

View .py

def custom_login(request, user, backend=None):
    """
    modificated generic.auth login.
    Send signal with extra parameter: previous [session_key]
    """

    # get previous seesion_key for signal 
    prev_session_key = request.session.session_key

    """ 
        original code

    """
    # send extra argument prev_session_key 
    user_logged_in.send(sender=user.__class__, request=request, user=user, prev_session_key=prev_session_key)

# custom class-based view overriden on LoginView
class CustomLoginView(LoginView):

    def form_valid(self, form):
        """Security check complete. Log the user in."""

        # changed default login
        custom_login(self.request, form.get_user())

        return HttpResponseRedirect(self.get_success_url())

当我基于默认的login()制作custom_login时，恐怕这不是最好的方法，因为我正在复制部分原始代码。也许在这里使用装饰器更好？