NameError: name 'JSONResponseMixin' is not defined

Question

我收到此错误，但我不明白为什么？我正在关注 Django 文档，部分错误代码来自 https://docs.djangoproject.com/en/3.0/topics/class-based-views/mixins/（link 中的最后一个代码）

错误：

/views.py", line 11, in <module>
    class ProgramListView(ListView, JSONResponseMixin, SingleObjectTemplateResponseMixin):
NameError: name 'JSONResponseMixin' is not defined

我在 views.py 中的代码：

from programs.models import Program
from django.views.generic import ListView, DetailView

from django.http import JsonResponse
from django.views.generic.detail import SingleObjectTemplateResponseMixin
from django.views.generic.detail import BaseDetailView




class ProgramListView(ListView, JSONResponseMixin, SingleObjectTemplateResponseMixin):
    
    model = Program
    template_name = 'programs/program_list.html'
    context_object_name = 'programs'
    paginate_by = 6
    paginate_orphans = 3

    def get_queryset(self):

        url_parameter = self.request.GET.get('q')

        # if self.request.method == 'GET' and self.request.is_ajax():
        if self.request.method == 'GET':
            queryset = Program.objects.filter(degree = url_parameter)
            return queryset

        if self.request.method == 'GET' and url_parameter is None:
            queryset = Program.objects.all()
            return queryset


    def render_to_response(self, context, url_parameter='BSc'):
        # Look for a 'format=json' GET argument
        # if self.request.GET.get('format') == 'json':
        # if self.request.is_ajax():
        if url_parameter=='BSc':
            return self.render_to_json_response(context)
        else:
            return super().render_to_response(context)

Answer 1

这是因为您没有导入 JSONResponseMixin。事实上 Django 没有这样的 mixin，这只是一个例子。

如果您真的想添加它，请在您的 class ProgramListView

上方添加此代码

from django.http import JsonResponse

class JSONResponseMixin:
    """
    A mixin that can be used to render a JSON response.
    """
    def render_to_json_response(self, context, **response_kwargs):
        """
        Returns a JSON response, transforming 'context' to make the payload.
        """
        return JsonResponse(
            self.get_data(context),
            **response_kwargs
        )

    def get_data(self, context):
        """
        Returns an object that will be serialized as JSON by json.dumps().
        """
        # Note: This is *EXTREMELY* naive; in reality, you'll need
        # to do much more complex handling to ensure that arbitrary
        # objects -- such as Django model instances or querysets
        # -- can be serialized as JSON.
        return context

来自https://docs.djangoproject.com/en/3.0/topics/class-based-views/mixins/#jsonresponsemixin-example

或者，您可以像这样直接 return 一个 JSON 响应：

if url_parameter=='BSc':
            return JsonResponse(context)

NameError: name 'JSONResponseMixin' is not defined

NameError: name 'JSONResponseMixin' is not defined

python

django

django-views

python-3.x

django-3.0