如何在graphene-django中使用django抽象类？

Question

我正在尝试为两个相似的具体类提供一个独特的接口，并从一个公共抽象类继承。

我的 Django 模型类：

class Metadata(models.Model):
    name = models.CharField(max_length=255)
    sequence = models.PositiveSmallIntegerField()
    is_choices = False

    class Meta:
        abstract = True


class MetadataScalar(Metadata):
    string_format = models.CharField(max_length=255, blank=True, null=True)


class MetadataChoices(Metadata):
    is_choices = True
    choices = models.CharField(max_length=255, blank=True, null=True)

我的石墨烯-django api：

class MetadataNode(DjangoObjectType):
    class Meta:
        interfaces = (Node,)
        connection_class = Connection
        model = Metadata
        fields = '__all__'


class MetadataScalarNode(MetadataNode):
    class Meta:
        interfaces = (Node,)
        connection_class = Connection
        model = MetadataScalar
        fields = '__all__'


class MetadataChoicesNode(MetadataNode):
    class Meta:
        interfaces = (Node,)
        connection_class = Connection
        model = MetadataChoices
        fields = '__all__'


class CreateMetadata(ClientIDMutation):
    metadata = Field(MetadataNode)

    class Input:
        name = String(max_length=255, required=True)
        sequence = Int(required=True)
        string_format = String()
        choices = List(String)

    @classmethod
    def mutate_and_get_payload(cls, root, info, **input):
        if 'string_format' in input:
            metadata = MetadataScalar.objects.create(
                name=input.get('name'),
                sequence=input.get('sequence'),
                string_format=input.get('string_format')
            )
        elif 'choices' in input:
            metadata = MetadataChoices.objects.create(
                name=input.get('name'),
                sequence=input.get('sequence'),
                choices=','.join(input.get('choices'))
            )
        return CreateMetadata(metadata=metadata)

当查询到CreateMetadata对应的 graphql 突变时，成功创建了元数据具体类。 ;-)

问题是，当查询在结果中要求创建具体的Metadata （这里是MetadataScalar或MetadataChoices ）时， graphql找不到具体类的节点并输出以下错误消息：

Expected value of type "MetadataNode" but got: MetadataScalar.

以下是一个示例查询，供您参考：

mutation {
  createMetadata (input: {
    stringFormat: "foo"
    sequence: 12
    name: "bar"
  }) {
    metadata {
      name
      sequence
    }
  }
}

如何让它很好地工作，而不必在查询的第二部分指定两种不同的结果类型（ metadataScalar和metadataChoices变量）？

Answer 1

您可以使用Union来指定多个不同的结果类。

在您的情况下，那将是：

class MetadataScalarNode(DjangoObjectType):
    class Meta:
        interfaces = (Node,)
        connection_class = Connection
        model = MetadataScalar
        fields = '__all__'


class MetadataChoicesNode(DjangoObjectType):
    class Meta:
        interfaces = (Node,)
        connection_class = Connection
        model = MetadataChoices
        fields = '__all__'


class MetadataNode(Union):
    class Meta:
        types = (MetadataScalarNode, MetadataChoicesNode)

graphql 查询将如下所示：

mutation {
  createMetadata (input: {
    stringFormat: "foo"
    sequence: 12
    name: "bar"
  }) {
    metadata {
      __typename
      ... on MetadataScalarNode {
        name
        sequence
        stringFormat
      }
      ... on MetadataChoicesNode {
        name
        sequence
        choices
      }
    }
  }
}

Answer 2

你试一试

... on metadata{
  name
  sequence
}

与您的界面。 Union 不能有任何字段，因此如果使用 Union 与 Interfaces，您需要有重复项。 https://docs.graphene-python.org/en/latest/types/unions/