Question

我正在尝试为“复杂”对象实现GrapgQL变异。假设我们有一个Contact，其中有三个字段：firstName，lastName和address，这是一个字段street的对象：

这是我的python方案实现：

class Address(graphene.ObjectType):
    street = graphene.String(description='Street Name')


class AddressInput(graphene.InputObjectType):
    street = graphene.String(description='Street Name', required=True)


class Contact(graphene.ObjectType):
    first_name = graphene.String(description='First Name')
    last_name = graphene.String(description='Last Name')
    address = graphene.Field(Address, description='Contact address')


class ContactInput(graphene.InputObjectType):
    first_name = graphene.String(description='First Name', required=True)
    last_name = graphene.String(description='Last Name', required=True)
    address = AddressInput(description='Contact address', required=True)


class CreateContact(graphene.Mutation):
    class Input:
        contact = ContactInput()

    contact = graphene.Field(Contact, description='Created Contact object')

    @staticmethod
    def mutate(instance, args, context, info):
        contact = Contact(**args['contact'])
        return CreateContact(contact=contact)

当我运行此查询时：

mutation CreateContact($contact: ContactInput!) {
    createContact(contact: $contact) {
        contact {
            firstName
            address {
                street
            }
        }
    }
}

使用以下变量：

{
    "contact": {
        "address": {
            "street": "Bond Blvd"
        },
        "firstName": "John",
        "lastName": "Doe"
    }
}

我得到以下结果：

{
    "createContact": {
        "contact": {
            "address": {
                "street": null
            },
            "firstName": "John"
        }
    }
}

如您所见，street字段在结果中为null。

如果我将mutate方法改为：

，我可以得到我需要的东西

@staticmethod
def mutate(instance, args, context, info):
    args['contact']['address'] = Address(**args['contact']['address'])
    contact = Contact(**args['contact'])
    return CreateContact(contact=contact)

但我不确定这是否正确。

因此，请建议正确的方法来启动嵌套结构。

嵌套对象的变异

0 个答案: