我如何在这里将两个模板视图合并到单个视图中？答案

【问题标题】：how do i combine two template views into on single view here?我如何在这里将两个模板视图合并到单个视图中？
【发布时间】：2022-01-03 14:08:56
【问题描述】：

我已经实现了两个视图来根据choice_fields显示数据，但是我有两个视图，视图和模板中的逻辑略有不同，我如何将它们组合成一个以便我处理DRY

views.py：

class View1(LoginRequiredMixin,TemplateView):
    template_name = 'temp1.html'
    
    def get_context_data(self, **kwargs):
        context =  super(View1,self).get_context_data(**kwargs)
        context['progress'] = self.kwargs.get('progress', 'in_progress')
        if context['progress'] == 'in_progress':
            context['objects'] = Model.objects.filter(progress='in_progress')
        else:
            context['objects'] = Model.objects.filter(progress__iexact=context['progress'], accepted=self.request.user)
        return context



class View2(LoginRequiredMixin,TemplateView):
    template_name = 'temp2.html'
    
    def get_context_data(self, **kwargs):
        context =  super(View2,self).get_context_data(**kwargs)
        context['progress'] = self.kwargs.get('progress', 'in_progress')
        if context['progress'] == 'in_progress':
            context['objects'] = Model.objects.filter(progress='in_progress',created = self.request.user)
        else:
            context['objects'] = Model.objects.filter(progress__iexact=context['progress'], created_by=self.request.user)
        return context

【问题讨论】：

您可以创建一个视图并覆盖 get_context_data 方法以返回不同的查询集，然后覆盖 get_template_names 方法以返回不同的模板

标签： django django-models django-views django-templates django-class-based-views

【解决方案1】：

在基于类的视图中实现 get_queryset 之类的东西。

class BaseView(LoginRequiredMixin,TemplateView):
    """ requires subclassing to define template_name and 
        update_qs( qs, progress) method """

    def get_context_data(self, **kwargs):
        context =  super(View1,self).get_context_data(**kwargs)
        progress = self.kwargs.get('progress', 'in_progress')

        if progress == 'in_progress':
            qs = Model.objects.filter(progress='in_progress')
        else:
            qs = Model.objects.filter(progress__iexact=context['progress'] )

        qs = self.update_qs( qs, (progress == 'in_progress') )

        context['progress'] = progress
        context['objects'] = qs
        return context

class View1( BaseView):
    template_name = 'temp1.html'

    def update_qs( self, qs, in_progress):
        if in_progress:
            return qs
        else:
            return qs.filter( accepted=self.request.user)

class View2( BaseView):
    template_name = 'temp2.html'

    def update_qs( self, qs, in_progress):
        if in_progress:
            return qs.filter( created = self.request.user)
        else:
            return qs.filter( created_by=self.request.user)

【讨论】：