Question

我正在尝试创建一个Django页面，其中可以更新某些内容，并且可以在分页表中查看某些内容。该模型如下所示：

class CostGroup(models.Model):
    name = models.CharField(max_length=200)
    description = models.CharField(max_length=200)

    def get_absolute_url(self):
        return reverse(
            'costgroup_detail', 
            kwargs={
                'costgroup_pk': self.pk,
            }
        )


class Cost(models.Model):
    cost_group = models.ForeignKey(CostGroup)
    amount = models.DecimalField(max_digits=50, decimal_places=2)

    def get_absolute_url(self):
        return reverse(
            'cost_detail', 
            kwargs={
                'cost_pk': self.pk,
            }
        )

因此，编辑表单适用于name模型的description和CostGroup字段，表格应显示“金额”列表

我以前只使用表单模板中包含的表单和表UpdateView来工作。现在，因为我想在表格中包含分页，我需要在同一页面上使用两个视图。我设计的页面最终应该看起来像这样：

我并不担心目前的风格，我目前的主要焦点是将表格和表格放在同一页面上。在目前的状态下，我唯一没有的就是桌子的分页：

目前的观点如下：

class CostDetail(UpdateView):
    model = models.Cost
    pk_url_kwarg = 'cost_pk'
    template_name = 'main/cost_detail.html'
    form_class = forms.CostDetailEditForm
    success_url = reverse_lazy('cost_list')

我有一种感觉，利用Django CBV使用的基础混合物可能是要走的路，但我不知道如何开始。

非常感谢任何帮助

感谢您的时间

Answer 1

（这个澄清似乎更好地作为新答案）

看起来你正在处理这两个表。对象级别使用CostGroup，而列表视图显示链接到CostGroup的成本中的子记录。假设这是真的，我将继续这样做：

class CostDetail(ModelFormMixin, ListView):

    model = CostGroup   # Using the model of the record to be updated
    form_class = YourFormName  # If this isn't declared, get_form_class() will
                               # generate a model form 


    ordering =  ['id']
    paginate_by = 10
    template_name = 'main/cost_detail.html' # Must be declared

    def get_queryset(self):
        # Set the queryset to use the Cost objects that match the selected CostGroup
        self.queryset = Cost.objects.filter(cost_group = get_object())
        # Use super to add the ordering needed for pagination
        return super(CostDetail,self).get_queryset()

    # We want to override get_object to avoid using the redefined get_queryset above
    def get_object(self,queryset=None):  
        queryset = CostGroup.objects.all()
        return super(CostDetail,self).get_object(queryset))

    # Include the setting of self.object in get()
    def get(self, request, *args, **kwargs):

        # from BaseUpdateView
        self.object = self.get_object()

        return super(CostDetail,self).get(request, *args, **kwargs)

    # Include the contexts from both
    def get_context_data(self, **kwargs):

        context = ModelFormMixin.get_context_data(**kwargs)
        context = ListView.get_context_data(**context)
        return context

    # This is the post method found in the Update View
    def post(self, request, *args, **kwargs):

        # From BaseUpdateView
        self.object = self.get_object()

        # From ProcessFormView
        form = self.get_form()
        self.form = form
        if form.is_valid():
            return self.form_valid(form)
        else:
            return self.form_invalid(form)

    def put(self, *args, **kwargs):
        return self.post(*args, **kwargs)

我没有尝试过这个，所以可能会有错误。祝你好运！

（请记住ccbv.co.uk是您在挖掘基于类的视图时的朋友）

Answer 2

我正在开发的应用程序现在使用类似的方法。我从ListView开始，引入FormMixin，然后从FormView引入post（）。

class LinkListView(FormMixin, ListView):
    model = Link
    ordering =  ['-created_on']
    paginate_by = 10
    template_name = 'links/link_list.html'

    form_class = OtherUserInputForm

    #=============================================================================#
    #
    #     Handle form input
    #

    def post(self, request, *args, **kwargs):
        """
        Handles POST requests, instantiating a form instance with the passed
        POST variables and then checked for validity.
        """
        form = self.get_form()
        self.form = form
        if form.is_valid():
            return self.form_valid(form)
        else:
            return self.form_invalid(form)

    def put(self, *args, **kwargs):
        return self.post(*args, **kwargs)

    def get_success_url(self):
        return reverse('links')

您可能还希望覆盖get_object()，get_queryset()和get_context()。

在同一模板页面上显示分页ListView和UpdateView

2 个答案: