Question

我正在使用django 1.8开发一个网站。这是其中一个视图的示例：

class ProfileView(View):
    template_name = 'index.html'

    # Return profile of any role (client/employee/admin)
    # Login required
    def get(self, request, *args, **kwargs):
        try:
            profile = Profile.objects.get(user=request.user)
            agency = None

            if request.user.is_employee():
                employee = EmployeeProfile.objects.get(profile=profile)
                agency = employee.agency

            if request.user.is_manager():
                agency = Agency.objects.get(manager=request.user)

        except (Profile.DoesNotExist, EmployeeProfile.DoesNotExist, Agency.DoesNotExist) as e:
            return HttpResponseRedirect('/404')

        return render(request, self.template_name, {"profile": profile, "agency": agency})

    # Client sign up
    # No decorator is needed
    def post(self, request):
        sign_up = SignUpForm(request.POST, request.FILES)

        response = json.loads(utils.create_user(sign_up,request.POST['avatar']))
        if response['profile'] is None:
            return JsonResponse({"code": 400, "response": response['message']})
        profile = serializers.deserialize("json", response['profile']).next().object

        group = Group.objects.get(name='clients')
        profile.user.groups.add(group)

        return HttpResponseRedirect('/')

问题是，我可以根据对控制器（视图）的请求类型设置不同的装饰器吗？

Answer 1

自Django 1.9 according to the doc起，可以通过以下方式应用装饰器：

from django.contrib.auth.decorators import login_required
from django.utils.decorators import method_decorator

@method_decorator(login_required, name='dispatch')
class YourClassBasedView(TemplateView):
    ...

其中name是要修饰的方法的名称。

或在装饰很少的情况下：

from django.contrib.auth.decorators import login_required
from django.views.decorators.cache import never_cache
from django.utils.decorators import method_decorator

decorators = [never_cache, login_required]

@method_decorator(decorators, name='dispatch')
class YourClassBasedView(TemplateView):
    ...

Answer 2

您需要将装饰器应用于基于类的视图的dispatch方法。这可以按如下方式完成：

class ProfileView(View):

    @youdecorator
    def dispatch(self,request,*args,**kwargs):
        return super(ProfileView,self).dispatch(request,*args,**kwargs)
    //Rest of your code.

Answer 3

有一个非常简单的解决方案可以达到你想要的效果，并不意味着装饰dispatch方法。您必须使用method_decorator覆盖您的方法（获取/发布）并将装饰器调用（不是装饰器本身）作为参数传递。

在你的情况下，它将是：

from django.utils.decorators import method_decorator

class ProfileView(View):
template_name = 'index.html'

    # Return profile of any role (client/employee/admin)
    @method_decorator(login_required())
    def get(self, request, *args, **kwargs):
        ...

    # Client sign up
    # No decorator is needed
    def post(self, request):
        ...

注意login_required装饰者的画面。

你可以传递任何面向函数的装饰器，甚至是自定义装饰器。例如：

def white_list_check():
    def decorator(func):
        def wrapper(request, *args, **kwargs):
            ip = request.META.get('REMOTE_ADDR', '0.0.0.0')
            if ip in WHITE_LIST:
                return func(request, *args, **kwargs)
            else:
                return HttpResponseForbidden()
        return wrapper
    return decorator

然后，再次：

class YourView(View):

    @method_decorator(white_list_check())
    def get(self, request):
        ...

Answer 4

可以在urls.py中使用像never_cache这样的装饰器而不是旧的方式：在views.py中

e.g。 never_cache装饰器：

旧样式views.py：

from django.views.decorators.cache import never_cache

@never_cache
def oldstyle_view(request):
    # and so on

在使用基于类的视图时，在urls.py中：

from django.views.decorators.cache import never_cache

urlpatterns = patterns('',
     (r'someurl/^$', never_cache(SomeClassBasedView.as_view())),
)

编辑2015年8月1日

注意：对于那些没有在views.py中定义完整视图的视图，这可以很方便，除此之外，装饰器也可以应用于视图中的depatch方法。

Answer 5

您可以覆盖dispatch方法并根据请求类型调用不同的装饰器：

from django.utils.decorators import method_decorator

class ProfileView(View):
    ... 

    def dispatch(self, *args, **kwargs):
        dispatch_method = super(ProfileView, self).dispatch

        if self.request.method == 'GET':
            dispatch_method = method_decorator(my_get_decorator)(dispatch_method)
        elif self.request.method == 'POST':
            dispatch_method = method_decorator(my_post_decorator)(dispatch_method)

        return dispatch_method(*args, **kwargs)

Django - 如何在基于类的视图方法中使用装饰器？

5 个答案: