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[英]How do I return the result of a class-based view from another class-based view in Django?
[英]Class-based view delegate to another view
是否可以將基於類的視圖委托給特定的基於類的視圖? 具體來說,我想做的是/指向一個名為'home'的視圖,如果用戶已登錄,則指向View A的主視圖;如果沒有用戶登錄,則指向View B.或者我可以執行重定向到其他網址。 我不確定這里最好的做法是什么。
您可以使用在網址中使用的相同方式從視圖中調用另一個視圖
class HomeView( TemplateView ):
template_name="index.html"
def dispatch( self, request, *args, **kwargs ):
if request.user.is_authenticated():
view=UserHomeView.as_view()
return view( request, *args, **kwargs )
return super( HomeView, self ).dispatch( request, *args, **kwargs )
class UserHomeView( TemplateView ):
template_name="user.html"
您只需重定向到另一個網址,該網址也可以通過基於類的視圖提供。
urls.py
url(r'^$', HomeView.as_view(), name='home'),
url(r'^login/', LoginView.as_view(), name='login'),
url(r'^welcome/$', WelcomeView.as_view(), name='welcome')
views.py
class HomeView(TemplateView):
def get(self, request, *args, **kwargs):
if request.user.is_authenticated():
return HttpResponseRedirect(reverse('welcome'))
else:
return HttpResponseRedirect(reverse('login'))
class WelcomeView(TemplateView):
def get(self, request, *args, **kwargs):
#do something
class LoginView(TemplateView):
def get(self, request, *args, **kwargs):
#show login page
確保用戶必須通過身份驗證的最佳做法是使用Mixin:
from django.contrib.auth.decorators import login_required
from django.utils.decorators import method_decorator
from django.views.generic import TemplateView
class LoginRequiredMixin(object):
u"""Ensures that user must be authenticated in order to access view."""
@method_decorator(login_required)
def dispatch(self, *args, **kwargs):
return super(LoginRequiredMixin, self).dispatch(*args, **kwargs)
class MyView(LoginRequiredMixin, TemplateView):
def get(self, request, *args, **kwargs):
#do something
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