Question

HttpResponseRedirect不起作用，它显示错误，因为“上下文必须是字典而不是响应”

class FooterLinksView(TemplateView):
    template_name = 'pages/footerlinks.html'
    model = FooterLink

    def get_context_data(self, **kwargs):

        context = super(FooterLinksView,self).get_context_data(**kwargs)
        string_name = self.kwargs['string']
        obj = FooterLink.objects.get(link_url=string_name)

        if obj.link_type == 'page':
            try:
                context['page_obj'] = obj
                return context
            except:
                pass
        else:
            pass
            print(obj.url_name)
            return HttpResponseRedirect(str(obj.url_name))

Answer 1

您必须始终从dict函数返回一个get_context_data。如果要重定向，也应该覆盖dispatch：

class FooterLinksView(TemplateView):
    ...
    def dispatch(self, request, *args, **kwargs):
        string_name = self.kwargs['string']
        obj = FooterLink.objects.get(link_url=string_name)
        if obj.link_type != 'page':
            return HttpResponseRedirect(str(obj.url_name))
        return super(FooterLinksView, self).dispatch(request, *args, **kwargs)

    def get_context_data(self, **kwargs):
        context = super(FooterLinksView,self).get_context_data(**kwargs)
        string_name = self.kwargs['string']
        obj = FooterLink.objects.get(link_url=string_name)
        context['page_obj'] = obj
        return context

Answer 2

get_context_data()是返回上下文的方法，因此它必须返回context对象。如果要重定向，请在get()中进行。

如下所示

FooterLinksView（TemplateView）类： template_name ='pages / footerlinks.html' 型号= FooterLink

def get(self, request, *args, **kwargs):
    string_name = self.kwargs['string']
    obj = FooterLink.objects.get(link_url=string_name)

    if obj.link_type != 'page':
        return HttpResponseRedirect(str(obj.url_name))
    else:
        return super().get(*args, **kwargs)
        # if python2
        # return super(FooterLinksView, self).get(*args, **kwargs)

def get_context_data(self, **kwargs):

    context = super(FooterLinksView,self).get_context_data(**kwargs)
    string_name = self.kwargs['string']
    obj = FooterLink.objects.get(link_url=string_name)

    if obj.link_type == 'page':
        try:
            context['page_obj'] = obj
            return context
        except:
            pass
    return context

或者您可以只在extra_context中使用TemplateView。

FooterLinksView（TemplateView）类： template_name ='pages / footerlinks.html' 型号= FooterLink

def get(self, request, *args, **kwargs):
    string_name = self.kwargs['string']
    obj = FooterLink.objects.get(link_url=string_name)

    if obj.link_type != 'page':
        try:
            self.extra_context = {
                'page_obj': obj
            }
        except:
            pass
        return HttpResponseRedirect(str(obj.url_name))
    else:
        return super().get(*args, **kwargs)
        # if python2
        # return super(FooterLinksView, self).get(*args, **kwargs)

Answer 3

get_context_data中的代码正在尝试检索单个对象并将其传递给上下文字典。此功能已经内置在DetailView中，并且似乎比TemplateView更合适。

from django.views.generic import DetailView

class FooterLinksView(DetailView):
    model = FooterLink
    template_name = 'pages/footerlinks.html'

    def get(self, request, *args, **kwargs):
        self.object = self.get_object()

        if self.object.link_type == 'page':
            context = self.get_context_data(object=self.object)
            return self.render_to_response(context)

        return HttpResponseRedirect(str(obj.url_name))

现在我们唯一需要添加的就是如何在FooterLink方法中检索get_object对象。如果您的link_url的{{1}}字段是主键，我们可以在类的顶部简单地添加FooterLink。否则，您应该使用pk_url_kwarg = 'link_url'和slug_field。

slug_url_kwarg

注意：您现在应该在模板中使用class FooterLinksView(DetailView): model = FooterLink pk = 'link_url' # if link_url is the primary key slug_field = 'link_url' # if link_url is not the primary key slug_url_kwarg = 'string' # if link_url is not the primary key template_name = 'pages/footerlinks.html' ...变量（而不是object）或覆盖另一个选项：obj。

Answer 4

class FooterLinksView(TemplateView):                                
    template_name = 'pages/footerlinks.html'                        
    model = FooterLink                                              

    def get(self, request, args, *kwargs):                        

        context = self.get_context_data(**kwargs)                   
        string_name = self.kwargs['string']                         
        obj = get_object_or_404(FooterLink, link_url=string_name)   

        if obj.link_type == 'page':                                 
           context['page_obj'] = obj                               

           return render(request, self.template_name, context)     
        else:                                                       
            a = obj.url_name                                        
            return HttpResponseRedirect(a)

HttpResponseRedirect在Django中不起作用

4 个答案: