我想覆盖login
__init__.py
中的django.contrib.auth
功能
我做了一些步骤:
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。
实际上我希望使用重叠contrib.auth
方法
login
答案 0 :(得分:0)
完成这种方法。
urls.py
# separate my custom login from django default auth
path('accounts/login/', CustomLoginView.as_view(), name='login'),
path('accounts/', include('django.contrib.auth.urls')),
views.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())
当我根据默认的login()创建custom_login时,我担心这不是最佳方法,因为我正在复制原始代码的一部分。也许在这里使用装饰器会更好吗?