Python '；dict'；对象没有属性'_meta&x27；

错误：“dict”对象没有属性“\u meta”这与我如何调用字典有关吗？我想把两个物体合并成一个。然后将其解析为json

这是回溯

def participant_specific_donation(self, participant):
    donations = Donation.objects.filter(participant = participant)

    helper = {}

    for donation in donations:
        ##helper['donation_date'] = donation.date
        helper['donation_amount'] = donation.amount
    return helper

def participant_specific(request, participant):
helper = RelayFunctions()
info = helper.participant_specific_donation(participant)
info1 = helper.participant_specific_milestone(participant)

data = [ 'participant_specific_donation' : info , 'participant_specific_milestone' : info1 ]
json_serializer = serializers.get_serializer("json")()
response = json_serializer.serialize(data, ensure_ascii=False)
return HttpResponse(response, mimetype="application/json")

---

我在代码片段中看到了两个问题

首先，这里的逻辑问题是：

Traceback:
 File "/home/vtrelayc/lib/python2.6/site-packages/django/core/handlers/base.py" in get_response
   111.                         response = callback(request, *callback_args, **callback_kwargs)
 File "/home/vtrelayc/projects/relay/relayapp/views.py" in participant_specific
   192.     response = json_serializer.serialize(data, ensure_ascii=False)
 File "/home/vtrelayc/lib/python2.6/site-packages/django/core/serializers/__init__.py" in serialize
  98.     s.serialize(queryset, **options)
File "/home/vtrelayc/lib/python2.6/site-packages/django/core/serializers/base.py" in serialize
  46.             concrete_model = obj._meta.concrete_model

Exception Type: AttributeError at /participants/specific/1/
Exception Value: 'dict' object has no attribute '_meta'

def participant_specific(request, participant):
helper = RelayFunctions()
info = helper.participant_specific_donation(participant)
info1 = helper.participant_specific_milestone(participant)
data = [ 'participant_specific_donation' : info , 'participant_specific_milestone' : info1 ]
json_serializer = serializers.get_serializer("json")()
response = json_serializer.serialize(data, ensure_ascii=False) ...
return HttpResponse(response, mimetype="application/json")

如果捐赠有多个元素，那么您将只得到列表中的最后一个元素，因为您正在循环

捐赠

，并且每次都将值分配给

帮助者

我认为您需要计算总金额，您可以按如下所示进行计算：

for donation in donations:
    ##helper['donation_date'] = donation.date
    helper['donation_amount'] = donation.amount

第二，线路：

helper['donation_amount'] += donation.amount

将给您一个语法错误

看来你这里需要一本字典。初始化词典的正确方法是使用大括号

{}

：

data = [ 
    'participant_specific_donation': info,
    'participant_specific_milestone': info1 
]

---

如果您希望所有捐款集中在一个地方，您可以先列出一个清单：

data = { 
    'participant_specific_donation': info, 
    'participant_specific_milestone': info1
}

然后将其指定给帮助器词典：

donation_amount = [donation for donation in donations]

---

data=[…

行将产生语法错误。这是真正的代码吗？data=['foo'：bar]看起来很奇怪。这是数组还是dict？能否发布准确的回溯，以及回溯引用位置周围的代码？t-8ch:data={'obj1'：obj1，'obj2'：obj2}json.dumps（数据）来自：谢谢你，AshRj！！制作一个字典是可行的，但我确实想要“捐赠”中的每个元素。所以我会在字典中制作一个数组来实现这一点？我知道这已经有2年了，但你的意思是

捐赠金额=[generation.amount for generation in generations]

？你写的清单理解除了复制一份

捐款

，没有任何作用。

helper['donation_amount'] = donation_amount