返回空字段的GraphQL嵌套查询

Question

我正在努力学习GraphQL（

query getLocationByPerson {
   People {
        firstName
        lastName
        service {
        location
      }
   }
}

我只得到一级字段，比如：

{
  "data": {
    "People": [
      {
        "firstName": "xxx",
        "lastName": "XXXXX",
        "service": null
      }
}

以下是我的模式：

type Query {
  People: [PeopleObject]!
  PeopleByName(lastName: String!): [PeopleObject]
  PeopleByID(id:ID!): [PeopleObject]
  Service: [ServiceObject]
  ServiceByID(id:ID!): [ServiceObject]
}
type PeopleObject {
    id: ID!
    Xid: String!
    firstName: String!
    lastName: String!
    email: String!
    serviceId: String
    apps: [String]
    service: [ServiceObject]
}
type ServiceObject {
    id: ID!
    name: String!
    location: String!
}

我的解析器：

const queries = {
    People: () => People.find({}),
    PeopleByName: (root,args,context,info) => People.find({lastName: args.lastName}),
    PeopleByID: (root,args,context,info) => People.find({_id: args.id}),
    Service: () => Service.find({}),
    ServiceByID: (root,args,context,info) => Service.find({_id: args.id})
  };

知道我做错了什么吗？

Answer 1

您的语法非常适合从graphql获取任何数据

type Query {
  People: [PeopleObject]!
}
type PeopleObject {
    id: ID!
    Xid: String!
    firstName: String!
    lastName: String!
    email: String!
    serviceId: String
    apps: [String]
    service: [ServiceObject]
}
type ServiceObject {
    id: ID!
    name: String!
    location: String!
}

为了简单起见

type Query {
  People: [{
    id: ID!
    Xid: String!
    firstName: String!
    lastName: String!
    email: String!
    serviceId: String
    apps: [String]
    service: [{
        id: ID!
        name: String!
        location: String!
    }]
}]!
}

但你只有人回来

People: () => People.find({}),

您应该返回服务数组对象，以便与人一起获取该返回对象

人：（）=

[{
    id: '5c5ac87e5e4d85ae77de9c50'
    Xid: 'xx'
    firstName: 'yyy'
    lastName: 'zzz'
    email: '[e-mail]'
    serviceId: '5c5ac87e5e4d85ae77de9c51'
    apps: ['testing', 'google']
    service: [{
        id: 5c5ac78d5e4d85ae77de9c4f,
        name: 'Tester',
        location: 'US America',
    }, {}]
}, {}]

如果您使用mongoose，请使用聚合从mongoBD获取数据。

Answer 2

您需要传递服务中location对象的引用，该引用与您的mongoose模型相同（我假设您使用的是mongoose）。传递引用后，可以添加填充（reference）。服务查询中的exec（）引用以下内容