小编给大家分享一下Django响应JsonResponse返回json格式数据报错怎么办,希望大家阅读完这篇文章后大所收获,下面让我们一起去探讨吧!

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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