架构

    class FirstType(DjangoObjectType):
        class Meta:
            model = FirstModel

    class SecondType(DjangoObjectType):
        class Meta:
            model = SecondModel

    class UnionType(graphene.Union):
        class Meta:
            types = (FirstType, SecondType)

因此，在这种模式下，我想用[pks]列表中的pk查询FirstType和SecondType的所有对象

    query {
        all_items(pks: [1,2,5,7]){
          ... on FirstType{
             pk,
             color, 
          }

          ... on SecondType{       
             pk,        
             size,
          }
        }
     }

来自FirstType的

PK通常不在SecondType中。

我尝试过以下方法

    def resolve_items(root, info, ids):
        queryset1 = FirstModel.objects.filter(id__in=pks)
        queryset2 = SecondModel.objects.filter(id__in=pks)
        return queryset1 | queryset2

但是它给出了一个错误：“无法在两个不同的基本模型上组合查询。”

我希望查询得到以下响应：

    { 'data':
        {'all_items':[
           {'pk': 1,
            'color': blue
           },
           {'pk': 2,
            'size': 50.0
           },
           ...
          ]}
     }

那么解析器应该是什么样子？

Answer 1

好的，所以我太专心于合并查询集，而我并没有注意到我可以简单地返回一个列表。

因此，以下是可以为我提供所需答案的解决方案：

label_values({__name__=~"metric1|metric2|metric3", service=~"abc.*xyz"}, service)

Answer 2

有关联合类型的石墨烯文档非常稀疏。这是如何正确执行操作的示例：

from graphene import ObjectType, Field, List, String, Int, Union

mock_data = {
    "episode": 3,
    "characters": [
        {
            "type": "Droid",
            "name": "R2-D2",
            "primaryFunction": "Astromech"
        },
        {
            "type": "Human",
            "name": "Luke Skywalker",
            "homePlanet": "Tatooine"
        },
        {
            "type": "Starship",
            "name": "Millennium Falcon",
            "length": 35
        }
    ]
}


class Human(ObjectType):
    name = String()
    homePlanet = String()


class Droid(ObjectType):
    name = String()
    primaryFunction = String()


class Starship(ObjectType):
    name = String()
    length = Int()


class Character(Union):
   class Meta:
       types = (Human, Droid, Starship)

   @classmethod
   def resolve_type(cls, instance, info):
      if instance["type"] == "Human":
         return Human
     if instance["type"] == "Droid":
         return Droid
      if instance["type"] == "Starship":
         return Starship


class RootQuery(ObjectType):
    result = Field(SearchResult)

    def resolve_result(_, info):
        return mock_data

然后，对于类似

的查询

  query Humans {
    result {
      episode
      characters {
        ... on Droid {
          name
        }
        ... on Starship {
          name
        }
        ... on Human {
          name
        }
      }
    }
  }

它返回正确的结果。

如何解析graphene.Union类型？

架构

2 个答案: