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解决Django响应JsonResponse返回json格式数据报错问题

程序员文章站 2022-04-16 22:25:22
代码return jsonresponse({"name": "tom"})报错:typeerror: in order to allow non-dict objects to be seriali...

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:

'in order to allow non-dict objects to be serialized set the ' 'safe parameter to false.'

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