相关疑难解决方法(0)

我想返回到在提交后调用 CreateView、UpdateView 和 DeleteView 的 DetailView(BuildingUnitDetail) 的 url。例如：http : //127.0.0.1 : 8000/unit/13/

我发现了其他几个关于重定向到先前视图的问题/答案，但我找不到任何适合我的解决方案。主要是因为我不明白他们。似乎它应该是一个直接的解决方案，我想多了。

有没有无痛的解决方案？

任何帮助都会非常感谢，在这已经超过 2 天了

#urls.py

from django.conf.urls import url
from . import views
from cdpapp.views import BuildingList, BuildingDetail, BuildingUnitDetail, CreateWorkOrder, EditWorkOrder, DeleteWorkOrder


urlpatterns = [
    url(r'^$', BuildingList.as_view(), name='index'),
    url(r'^building/(?P<pk>\d+)/$', BuildingDetail.as_view(), name='building_detail'),
    url(r'^unit/(?P<pk>\d+)/$', BuildingUnitDetail.as_view(), name='building_unit_detail'),
    url(r'^workorder/add/$', CreateWorkOrder.as_view(), name='workorder_add'),
    url(r'^workorder/(?P<pk>\d+)/$', EditWorkOrder.as_view(), name='workorder_update'),
    url(r'^workorder/(?P<pk>\d+)/delete/$', DeleteWorkOrder.as_view(), name='workorder_delete'),
]

#views.py

class BuildingUnitDetail(DetailView):
    model = Unit
    template_name = 'cdpapp/building_units_detail.html'
    context_object_name = 'units'


class CreateWorkOrder(CreateView):
    template_name = 'cdpapp/workorder_form.html'
    model = WorkOrder
    success_url = …