Question

我尝试使用Django rest framework

构建API

我得到了

的对象

type 'TypeError' is not JSON serializable

我该怎么办？

这是我的view.py

class NewsViewSet(viewsets.ModelViewSet):
    queryset = News.objects.all()
    serializer_class = NewsSerializer

    def list(self, request, **kwargs):
        try:
            nba = query_nba_by_args(**request.query_params)
            serializer = NewsSerializer(nba['items'], many=True)
            result = dict()
            result['data'] = serializer.data
            result['draw'] = nba['draw']
            result['recordsTotal'] = nba['total']
            result['recordsFiltered'] = nba['count']
            return Response(result, status=status.HTTP_200_OK, template_name=None, content_type=None)

        except Exception as e:
            return Response(e, status=status.HTTP_404_NOT_FOUND, template_name=None, content_type=None)

Answer 1

Django无法将Exception对象转换为JSON格式并引发错误。要修复它，您应该将错误转换为字符串并将结果传递给响应：

except Exception as e:
    return Response(str(e), status=status.HTTP_404_NOT_FOUND, template_name=None, content_type=None)

Answer 2

如下所示更改您的观点，

class NewsViewSet(viewsets.ModelViewSet):
    queryset = News.objects.all()
    serializer_class = NewsSerializer

    def list(self, request, **kwargs):
        try:
            nba = query_nba_by_args(**request.query_params)
            serializer = NewsSerializer(nba['items'], many=True)
            result = dict()
            result['data'] = serializer.data
            result['draw'] = nba['draw'] if "draw" in nba else None
            result['recordsTotal'] = nba['total'] if "total" in nba else None
            result['recordsFiltered'] = nba['count'] if "count" in nba else None
            return Response(result, status=status.HTTP_200_OK)

        except Exception as e:
            return Response(str(e), status=status.HTTP_404_NOT_FOUND)

Answer 3

首先

import json
from django.http import HttpResponse

更改行

  return Response(result, status=status.HTTP_200_OK, template_name=None, content_type=None)

这个

   return HttpResponse(json.dumps(result),content_type="application/json")

或使用

 from django.http import JsonResponse

 return JsonResponse(json.dumps(result))

Answer 4

Python 异常不是 json 可序列化的。由于某些连接或内容不可用问题，它在 try 中失败，然后进入 except 块，在那里您将异常 e 原样传递给 Response()，从而创建问题。解决方案 - 检查 URL 并在 except 块中将异常 e 转换为字符串并传递给 Response(str(e), status=status.HTTP_404_NOT_FOUND, template_name=None, content_type=None)。

“TypeError”类型的对象不是JSON可序列化的

4 个答案: