Question

我有这个ListView，我正尝试从数据库中获取12辆汽车，并按价格对其进行排序，我在数据库中有20个条目，在分页按钮中它显示6个按钮，也就是6页，因此正在从db返回所有可能的数据，这时我迷路了，找不到错误，对于分页和排序都将非常感谢。

class CardListView(ListView):
    model = Car
    template_name = 'home.html'
    context_object_name = 'cars'
    ordering = ['-price']
    paginate_by = 5
    limit = 12

    def get_context_data(self, **kwargs):
        context = super().get_context_data(**kwargs)
        paginator = context['paginator']
        page_numbers_range = 10 
        max_index = len(paginator.page_range)

        page = self.request.GET.get('page')
        current_page = int(page) if page else 1

        start_index = int((current_page - 1) / page_numbers_range) * page_numbers_range
        end_index = start_index + page_numbers_range
        if end_index >= max_index:
            end_index = max_index

        page_range = paginator.page_range[start_index:end_index]
        cars = Car.objects.all()[:self.limit]
        car_types = CarCategory.objects.all()

        context['cars'] = cars

        context['car_types'] = car_types
        context['page_range'] = page_range
        return context

Answer 1

您无需自己执行分页逻辑，只需在get_context_data中实现要传递给模板的 additional 元素即可。分页逻辑由ListView itself处理。如果要限制查询集，则应在get_queryset(..) [Django-doc]函数中做到这一点：

class CardListView(ListView):
    model = Car
    template_name = 'home.html'
    context_object_name = 'cars'
    ordering = ['-price']
    paginate_by = 5
    limit = 12

    def get_queryset(self):
        return super().get_queryset()[:self.limit]

    def get_context_data(self, **kwargs):
        context = super().get_context_data()  # with paginated queryset
        context['car_types'] = car_types = CarCategory.objects.all()
        return context

原始的ListView实现将调用get_queryset，然后将其传递给适当的分页器。通过使用super()，我们还不需要关心ordering等。这些都由ListView本身处理。

Django：Paginator从数据库返回所有数据，并且排序不起作用

1 个答案: