Question

我的目标

在表格中列出我所有更新（模型）的网站
不要一次显示所有模型（分页 - 每页10个）
过滤并对列表进行排序

我的想法

我可以使用ListView获取一组所有更新
使用paginate_by = 10
使用表单在我的QuerySet中设置order_by或过滤

我的问题

我不知道如何使用过滤器和排序添加表单来修改我的QuerySet。我的想法是使用额外的filter和order_by修改get_queryset中的Query。

我的观点

class MyView(ListView):
    model = Update
    template_name = "updates/update.html"
    paginate_by = 10

    def get_queryset(self):
        return Update.objects.filter(
            ~Q(state=Update.STATE_REJECTED),
            ~Q(state=Update.STATE_CANCELED),
            ~Q(state=Update.STATE_FINISHED),
        ).order_by(
            'planned_release_date'
        )

我的想法

像这样的东西。我知道这不是这样的......只是为了说明

class MyView(ListView):
    model = Update
    template_name = "updates/update.html"
    paginate_by = 10

    def post(self, request, *args, **kwargs):
        new_context = Update.objects.filter(
            request.POST.get("filter"),
        ).order_by(
            request.POST.get("sorting"),
        )

    def get_queryset(self):
        return Update.objects.filter(
            ~Q(state=Update.STATE_REJECTED),
            ~Q(state=Update.STATE_CANCELED),
            ~Q(state=Update.STATE_FINISHED),
        ).order_by(
            'planned_release_date'
        )

Answer 1

你不需要发帖。在网址中传递过滤器值和order_by，例如：

.../update/list/?filter=filter-val&orderby=order-val

并在get_queryset中获取过滤器和orderby，如：

class MyView(ListView):
    model = Update
    template_name = "updates/update.html"
    paginate_by = 10

    def get_queryset(self):
        filter_val = self.request.GET.get('filter', 'give-default-value')
        order = self.request.GET.get('orderby', 'give-default-value')
        new_context = Update.objects.filter(
            state=filter_val,
        ).order_by(order)
        return new_context

    def get_context_data(self, **kwargs):
        context = super(MyView, self).get_context_data(**kwargs)
        context['filter'] = self.request.GET.get('filter', 'give-default-value')
        context['orderby'] = self.request.GET.get('orderby', 'give-default-value')
        return context

确保您提供正确的默认值以进行过滤和排序

示例表单（您可以根据需要进行修改）：

<form method="get" action="{% url 'update-list' %}">
    <p>Filter: <input type="text" value={{filter}} name="filter"/></p>
    <p>order_by: <input type="text" value={{orderby}} name="orderby"/></p>
    <p><input type="submit" name="submit" value="submit"/></p>
</form>

Answer 2

我想知道为什么没有人在这里提到这个很酷的库：django-filter https://github.com/carltongibson/django-filter

你可以定义你的过滤非常干净的逻辑，并获得快速工作的表格等。

演示：https://stackoverflow.com/a/46492378/953553

Answer 3

我在其他地方发布了这个，但我认为这会增加选定的答案。

我认为你最好通过get_context_data这样做。手动创建HTML表单并使用GET检索此数据。我写的一个例子如下。提交表单时，可以使用get数据通过上下文数据传回。此示例不适合您的请求，但应该有助于其他用户。

def get_context_data(self, **kwargs):
    context = super(Search, self).get_context_data(**kwargs)
    filter_set = Gauges.objects.all()
    if self.request.GET.get('gauge_id'):
        gauge_id = self.request.GET.get('gauge_id')
        filter_set = filter_set.filter(gauge_id=gauge_id)

    if self.request.GET.get('type'):
        type = self.request.GET.get('type')
        filter_set = filter_set.filter(type=type)

    if self.request.GET.get('location'):
        location = self.request.GET.get('location')
        filter_set = filter_set.filter(location=location)

    if self.request.GET.get('calibrator'):
        calibrator = self.request.GET.get('calibrator')
        filter_set = filter_set.filter(calibrator=calibrator)

    if self.request.GET.get('next_cal_date'):
        next_cal_date = self.request.GET.get('next_cal_date')
        filter_set = filter_set.filter(next_cal_date__lte=next_cal_date)

    context['gauges'] = filter_set
    context['title'] = "Gauges "
    context['types'] = Gauge_Types.objects.all()
    context['locations'] = Locations.objects.all()
    context['calibrators'] = Calibrator.objects.all()
    # And so on for more models
    return context

Answer 4

这是我们的工作方式，也使您可以进行验证/类型转换：

class UnitList(PermissionRequiredMixin, ListView):
    """ Class based view to show a list of all buildings for a specific user """

    model = Unit
    ordering = ['building', 'unit']
    paginate_by = 100

    # Access
    permission_required = ['core.manager_perm']
    raise_exception = True  # If true, give access denied message rather than redirecting to login

    def get_queryset(self):
        try:
            units = self.model.objects.filter(building__company=self.request.user.profile.company)
        except Profile.DoesNotExist:
            units = self.model.objects.none()

        form = UnitSearchForm(self.request.GET)

        if form.is_valid():
            filters = {}

            address = form.cleaned_data['address']
            neighborhood = form.cleaned_data['neighborhood']
            beds = form.cleaned_data['beds']
            amenity = form.cleaned_data['amenity']

            if address:
                filters['building__street_index__istartswith'] = compute_street_address_index(address)

            if neighborhood:
                filters['building__neighborhood__icontains'] = neighborhood

            if beds:
                filters['beds'] = beds

            if amenity:
                filters['unit_amenities__name__iexact'] = amenity
    
            units = units.filter(**filters)

        return units.select_related('building').order_by(*self.ordering)

    def get_context_data(self, **kwargs):
        context = super().get_context_data(**kwargs)
        context['form'] = UnitSearchForm(self.request.GET)

        return context

Django ListView - 用于过滤和排序的表单

4 个答案: