django 在 CreateView 之后重定向到上一个 url

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

我发现了有关重定向到先前视图的其他几个问题/答案，但我无法找到任何适合我的解决方案。主要是因为我不理解他们。看来这应该是一个直接的解决方案，而我想得太多了。

有没有无痛的解决方案？

任何帮助将不胜感激，已经这样做超过 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 = reverse_lazy('back to calling url')


class EditWorkOrder(UpdateView):
    template_name = 'cdpapp/workorder_form.html'
    model = WorkOrder
    success_url = reverse_lazy('back to calling url')


class DeleteWorkOrder(DeleteView):
    template_name = 'cdpapp/workorder_form.html'
    model = WorkOrder
    success_url = reverse_lazy('back to calling url')

#forms.py

class WorkOrderForm(forms.Form):
    building = forms.ModelChoiceField(queryset=Building.objects.all())
    unit = forms.ModelChoiceField(queryset=Unit.objects.all())
    ...

#表单模板

{% block content %}
<form method="POST"> {% csrf_token %}
    {{ form.as_p }}
    <input class="btn btn-danger" type="submit" value="Submit">
</form>
{% endblock content %}

使用下一个参数可能是一个优雅的解决方案。

这是一个示例（我刚刚编写的未经测试的代码）。基本上使用get_form_kwargs方法确保您的按钮参数在 GET 上被推送到表单的初始字典。这form_valid方法扩展确保success_url当 POST 数据中可用时，属性被重载

这样您仍然可以使用定义默认值success_url以普通方式拥有财产。

注意：您不能信任用户的输入。为了简单起见，我只是使用 CharField 作为下一个字段。在现实生活中，您应该检查来自该字段的数据并对其进行验证。

BuildingUnitDetail 模板

<a href="{% url 'workorder_add' %}?next={% url 'building_unit_detail' object.pk %}">
    Add workorder
</a>  <!-- assuming 'object' (Unit) is available in your template's context -->

工单表格模型

class WorkOrderForm(forms.ModelForm):
    next = forms.CharField(required=False)

    class Meta:
        model = WorkOrder
        exclude = tuple()

创建工作订单视图

class CreateWorkOrder(CreateView):
    template_name = 'cdpapp/workorder_form.html'
    form_class = WorkOrderForm
    
    def get_form_kwargs(self, **kwargs):
        kwargs = super(CreateWorkOrder, self).get_form_kwargs()
        redirect = self.request.GET.get('next')
        if redirect:
            if 'initial' in kwargs.keys():
                kwargs['initial'].update({'next': redirect})
            else:
                kwargs['initial'] = {'next': redirect}
        return kwargs

    def form_invalid(self, form):
        import pdb;pdb.set_trace()  # debug example
        # inspect the errors by typing the variable form.errors
        # in your command line debugger. See the pdb package for
        # more useful keystrokes
        return super(CreateWorkOrder, self).form_invalid(form)

    def form_valid(self, form):
        redirect = form.cleaned_data.get('next')
        if redirect:
            self.success_url = redirect
        return super(CreateWorkOrder, self).form_valid(form)
        
        

第二个想法？

避免重定向，您还可以在弹出窗口中处理这些“任务”或在单个视图中处理多个表单。这会增加复杂性，但后者might增强用户体验。