Question

在处理表单提交时，我必须执行自定义逻辑（联系Web服务等）。如果失败，我想阻止创建模型Car的新实例。

让我们通过一个简单的片段演示：

from django.views import generic
from django.http import HttpResponseRedirect

class CarCreateView(generic.edit.CreateView):
      model = Car
      form_class = CarForm

      def form_valid(self, form):

           # some logic that may succeed or fail

           if success:
               messages.success(self.request, 'bla bla') 
               return super().form_valid(form)
           else:
               messages.error(self.request, 'blabla')
               # How to return to form index without saving???
               return HttpResponseRedirect(self.get_success_url())

不调用super().form_valid(form)是不够的。一辆新车仍在保存中。有什么想法吗？

Answer 1

其实我一直都错了。愚蠢的错误。当我保存新实例时，执行从未达到过这一点。

对于任何寻找相同问题的人来说，更清晰的处理方法似乎是

return self.render_to_response(self.get_context_data(form=form))。

因此代码看起来像：

from django.views.generic import CreateView
from django.http import HttpResponseRedirect

class CarCreateView(CreateView):
      model = Car
      form_class = CarForm

      def form_valid(self, form):

           # some logic that may succeed or fail

           if success:
               messages.success(self.request, 'bla bla') 
               return super(CarCreateView, self).form_valid(form)
           else:
               messages.error(self.request, 'blabla')
               return self.render_to_response(self.get_context_data(form=form))

这样我们就可以返回相同的表单页面，而无需创建新的实例。

Answer 2

from django.views.generic import CreateView
from django.http import HttpResponseRedirect

class CarCreateView(CreateView):
      model = Car
      form_class = CarForm

      def form_valid(self, form):

           # some logic that may succeed or fail

           if success:
               messages.success(self.request, 'bla bla') 
               return super(CarCreateView, self).form_valid(form)
           else:
               messages.error(self.request, 'blabla')
               return super(CarCreateView, self).form_invalid(form)

从CreateView的form_valid（）返回而不保存实例

2 个答案: