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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