Django从另一个基于类的视图调用基于类的视图

psy*_*ok7 21 python django django-class-based-views

我试图调用基于类的视图,我能够做到这一点,但由于某种原因,我没有得到我正在调用的新类的上下文

class ShowAppsView(LoginRequiredMixin, CurrentUserIdMixin, TemplateView):
    template_name = "accounts/thing.html"



    @method_decorator(csrf_exempt)
    def dispatch(self, *args, **kwargs):
        return super(ShowAppsView, self).dispatch(*args, **kwargs)

    def get(self, request, username, **kwargs):
        u = get_object_or_404(User, pk=self.current_user_id(request))

        if u.username == username:
            cities_list=City.objects.filter(user_id__exact=self.current_user_id(request)).order_by('-kms')
            allcategories = Category.objects.all()
            allcities = City.objects.all()
            rating_list = Rating.objects.filter(user=u)
            totalMiles = 0
            for city in cities_list:
                totalMiles = totalMiles + city.kms

        return self.render_to_response({'totalMiles': totalMiles , 'cities_list':cities_list,'rating_list':rating_list,'allcities' : allcities, 'allcategories':allcategories})


class ManageAppView(LoginRequiredMixin, CheckTokenMixin, CurrentUserIdMixin,TemplateView):
    template_name = "accounts/thing.html"

    def compute_context(self, request, username):
        #some logic here                        
        if u.username == username:
            if request.GET.get('action') == 'delete':
                #some logic here and then:
                ShowAppsView.as_view()(request,username)
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我做错了什么人?

psy*_*ok7 42

代替

ShowAppsView.as_view()(self.request)
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我不得不这样做

return ShowAppsView.as_view()(self.request)
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  • 我发现如果你做ShowAppsView.as_view()(request,*args,**kwargs),实际上可以通过ContextMixin将args和kwargs传递给get_context_data方法,它们显示为self.args和self. kwargs.这对于重写此方法并添加到表单的上下文非常有用. (5认同)
  • 但是当使用 DRF 时,它会抛出```request` 参数必须是 `django.http.HttpRequest` 的实例,而不是 `rest_framework.request.Request`。``` (4认同)