解决Django响应JsonResponse返回json格式数据报错问题

脚本专栏 2024/12/25 佚名

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代码

return JsonResponse({"name": "tom"})

报错：

TYPEERROR: In order to allow non-dict objects to be serialized

set the safe parmeter to False

解决：

return JsonResponse({"name": "tom"}, safe=False)

增加safe=false，使其接受列表

补充知识：python 里面 JsonResponse (book_list,safe=False)

代码为：

# 查询所有图书 、 增加图书
def get(self,request):

  queryset = BookInfo.objects.all()
  book_list = []

  for book in queryset:
    book_list.append({
      'id':book.id,
      'bread':book.bread

    })
  return JsonResponse (book_list,safe=False)

遇到问题：

JsonResponse (book_list,safe=False)

safe=False 这是什么鬼？

解决方案：

down 下源码后：

def __init__(self, data, encoder=DjangoJSONEncoder, safe=True,
       json_dumps_params=None, **kwargs):
  if safe and not isinstance(data, dict):
    raise TypeError(
      'In order to allow non-dict objects to be serialized set the '
      'safe parameter to False.'
    )
 
 if json_dumps_params is None:
    json_dumps_params = {}
  kwargs.setdefault('content_type', 'application/json')
  data = json.dumps(data, cls=encoder, **json_dumps_params)
  super(JsonResponse, self).__init__(content=data, **kwargs)

最终答案：