结合DjangoObjectType和ObjectType

时间:2018-01-15 17:01:54

标签: django graphql graphene-python

我有一个带有计算属性字段clicks的简单django模型。该模型如下所示:

class Link(models.Model):
    url = models.URLField()

    @property
    def clicks(self):
        """
        Property does some calculations and returns a list of dictionaries:

        """
        # removed calculation for simplicity
        return [{'dt': 1, 'clicks': 100}, {'dt': 2, 'clicks': 201}] 

我想在我的graphql端点中访问此模型。所以我创建了以下类型和查询:

class Stats(graphene.ObjectType):
    clicks = graphene.String()
    dt = graphene.String()


class LinkType(DjangoObjectType):
    clicks = graphene.List(Stats, source='clicks')

    class Meta:
        model = Link


class Query(object):
    link = graphene.Field(LinkType, id=graphene.Int())

    def resolve_link(self, info, **kwargs):
        id = kwargs.get('id')
        url = kwargs.get('url')
        if id is not None:
            return Link.objects.get(pk=id)
        return None

现在我应该可以在graphql explorer中使用以下查询:

{
  link(id: 3) {
    id,
    url,
    clicks{
      clicks,
      dt
    }
  }
}

我的预期结果是这样的:

{
  id: 3,
  url: "www.google.de",
  clicks: [
    dt: 1, clicks: 100},
    dt: 2, clicks: 201}
  ]
}

clicksdt的嵌套值为null

{
  id: 3,
  url: "www.google.de",
  clicks: [
    dt: null, clicks: null},
    dt: null, clicks: null}
  ]
}

那我在这里做错了什么?如何在石墨烯中将dicts列表转换为ObjectType?

编辑:

我使用@ mark-chackerian的修改版本来解决问题: 好像我期待太多"魔术"来自石墨烯,我必须明确告诉它每个领域是如何解决的。

class Stats(graphene.ObjectType):
    clicks = graphene.String()
    dt = graphene.String()

    def resolve_clicks(self, info):
        return self['clicks']

    def resolve_dt(self, info):
        return self['dt']

1 个答案:

答案 0 :(得分:1)

您必须更明确地告诉石墨烯如何制作Stats对象列表。

尝试这样的事情:

class LinkType(DjangoObjectType):
    clicks = graphene.List(Stats)

    class Meta:
        model = Link

    def resolve_clicks(self, info):
        return [Stats(dt=click_dict['dt'], clicks=click_dict['clicks') for click_dict in self.clicks]