我如何防止CreateView（Django 1.6）将我重定向到success_url，如果该表单有效，那么只需返回一个json？

Question

我尝试覆盖返回json而不是HttpResponseRedirect的form_valid方法，但这不起作用。 请向我提出以下错误： "The view did not return an HttpResponse object."

这是我认为的代码：

from django.views.generic import CreateView
from django.shortcuts import render
from django.http import HttpResponse
from .forms import TestForm
import json,urllib2

class CreateViewTest(CreateView):
    form_class=TestForm
    template_name="Test.html"
    success_url = ""

    def form_valid(self, form):
        urllib2.urlopen('returnjson/',json.dumps(form.data))
        #returning here try different things but all gave me the same error

Answer 1

试试这个吧。

import json


class CreateViewTest(CreateView):
    form_class = TestForm
    template_name = 'test.html'

    def form_valid(self, form):
        # return an HttpResponse object
        return HttpResponse(json.dumps(form.data), content_type="application/json")

Answer 2

import json
from django.http import HttpResponse

def form_valid(self, form):
    return HttpResponse(json.dumps(form.data), content_type='application/json')

另外，出于可重用性的考虑，我建议您创建一个mixin ，仅在请求为AJAX时才负责呈现JSON响应：

from django.views.generic.edit import FormMixin

class JsonResponseFormMixin(FormMixin):
    def form_valid(self, form):
        if self.request.is_ajax():
            return HttpResponse(json.dumps(form.data), content_type='application/json')
        return super(JsonResponseFormMixin, self).form_valid(form)

    def form_invalid(self, form):
        if self.request.is_ajax():
            return return HttpResponse(json.dumps(form.errors), content_type='application/json', status_code=400)
        return super(JsonResponseFormMixin, self).form_valid(form)


from django.views.generic import CreateView
from .forms import TestForm

class CreateViewTest(CreateView, JsonResponseFormMixin):
    form_class = TestForm