多模型通用ListView转模板

Multiple models generic ListView to template

将 2 个模型列在通用 IndexView 中的最简单方法是什么?我的两个模型是 CharacterSeriesCharacterUniverse

我的views.py

    from .models import CharacterSeries, CharacterUniverse


    class IndexView(generic.ListView):
        template_name = 'character/index.html'
        context_object_name = 'character_series_list'

        def get_queryset(self):
            return CharacterSeries.objects.order_by('name')


    class IndexView(generic.ListView):
        template_name = 'character/index.html'
        context_object_name = 'character_universe_list'

        def get_queryset(self):
            return CharacterUniverse.objects.order_by('name')

我需要知道最短最优雅的代码。看了很多但不想使用mixins。我可能没有被指向正确的方向。

谢谢大家

您可以像这样在 ListView 上将一个查询集作为上下文传递,

class IndexView(generic.ListView):
    template_name = 'character/index.html'
    context_object_name = 'character_series_list'
    model = CharacterSeries

    def get_context_data(self, **kwargs):
        context = super(IndexView, self).get_context_data(**kwargs)
        context.update({
            'character_universe_list': CharacterUniverse.objects.order_by('name'),
            'more_context': Model.objects.all(),
        })
        return context

    def get_queryset(self):
        return CharacterSeries.objects.order_by('name')

https://docs.djangoproject.com/en/1.8/ref/class-based-views/mixins-simple/#django.views.generic.base.ContextMixin.get_context_data

听起来像 mixin 是唯一的(对)?方法。我添加了一个 get_context_data 方法,现在可以使用了。

快速提问,添加超过 2 个模型如何?

下面现在适用于 2 个模型:

class IndexView(generic.ListView):
    template_name = 'character/index.html'
    context_object_name = 'character_series_list'

    def get_queryset(self):
        return CharacterSeries.objects.order_by('name')

    def get_context_data(self, **kwargs):
        context = super(IndexView, self).get_context_data(**kwargs)
        context['character_universe_list'] = CharacterUniverse.objects.order_by('name')
        return context