我正在使用通用视图DetailView
和ListView
我有三个模型,例如User
,Business
和Invoice
。一个用户可以有多个业务,可以有多个发票。
#mixins.py
class BusinessOwnerRequiredMixin(object):
def has_permissions(self):
obj = self.get_object()
if isinstance(obj, Business):
# Assumes that your Article model has a foreign key called `auteur`.
return obj.owner == self.request.user
def dispatch(self, request, *args, **kwargs):
if not self.has_permissions():
raise PermissionDenied
return super(BusinessOwnerRequiredMixin, self).dispatch(request, *args, **kwargs)
#views.py
class BusinessDashboard(BusinessOwnerRequiredMixin, DetailView):
model = Business
template_name = "business/business-main.html"
class InvoiceListView(BusinessDashboard):
template_name = "business/purchase/purchase_invoice-main.html"
class InvoiceDetailView(InvoiceListView):
template_name = "business/purchase/purchase_invoice.html"
#urls.py
path(r'business/<pk>/purchase_invoices/<pid>/',vw.PurchaseInvoiceDetailView.as_view(), name='purchase_invoice'),
path(r'business/<pk>/purchase_invoices/',vw.PurchaseInvoiceListView.as_view(), name='purchase_invoices')
我要寻找的是从Business的DetailView继承发票的ListView,即从业务的实例继承,即必须列出特定业务的所有发票。
如何像这样实现:
#views.py
#views.py
class BusinessDashboard(BusinessOwnerRequiredMixin, DetailView):
model = Business
template_name = "business/business-main.html"
class InvoiceListView(BusinessDashboard, ListView):
model = Invoice
template_name = "business/purchase/purchase_invoice-main.html"
class InvoiceDetailView(InvoiceListView, DetailView):
model = Invoice
template_name = "business/purchase/purchase_invoice.html"
但是这是行不通的,因为我在每个课程上都覆盖了model
...
对于网址http://example.com/business/1/invoices/1/
,在模板内部,我必须有一个带有发票实例的变量。
答案 0 :(得分:2)
没有继承的必要;您只需要定义get_queryset
即可根据业务pk过滤发票。
class InvoiceListView(ListView):
template_name = "business/purchase/purchase_invoice-main.html"
def get_queryset(self):
return Invoice.objects.filter(business_id=self.kwargs['pk'])