Django中基于类的通用视图的问题

vim*_*vim 13 python django django-generic-views

我正在尝试使用基于Djangos类的通用视图编写CRUD应用程序.以下是我编写的用于在db中创建新用户的代码.

  from django.views.generic import CreateView
  from django.contrib.auth.decorators import login_required
  from django.contrib import messages

  class UserCreateView(CreateView):
  """ 
  Display and accept a new user to be created in db
  """
    form_class = ProfileForm
    template_name = 'userdb/profile_form.html'
    success_url = '/organization/users/'

    def post(self, request, *args, **kwargs):
      messages.success(request, "Success", extra_tags='msg')
      return super(UserCreateView, self).post(request, *args, **kwargs)

    @method_decorator(login_required)
    def dispatch(self, *args, **kwargs):
      return super(UserCreateView, self).dispatch(*args, **kwargs)
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请注意,要添加要显示给用户的成功消息,我必须扩展post函数.我知道这不是一个好方法,因为当调用此函数时,不确定提交的表单是否包含有效数据.所以我的问题是,是否有推荐的方法将Djangos消息传递框架与基于类的通用视图相结合?

Yuj*_*ita 16

答案取决于您对消息传递框架的具体要求.如果需要为每个get请求调用它,你自然需要将它放在get方法中(指出没有人放置这个代码的正确位置).

无论如何,听起来你正在寻找一个只在表格有效时触发的地方.

CreateView使用the ModelFormMixin实现一个form_valid方法,该方法仅在成功保存表单时触发.完善!

def form_valid(self, form):
    messages.success(self.request, "Success", extra_tags='msg')
    return super(UserCreateView, self).form_valid(form)  
    # ModelFormMixin will now save
    # FormMixin will now redirect to success_url()
    # override above behavior if you need to do something with the object
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