GraphQL 嵌套查询返回空字段答案

【问题标题】：GraphQL nested query returning null fieldGraphQL 嵌套查询返回空字段
【发布时间】：2019-07-08 08:44:48
【问题描述】：

我正在尝试学习 GraphQL（& node.js & MongoDB 等）。我无法让这个简单的嵌套查询返回结果：

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

我只得到一级字段，例如：

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

这是我的架构：

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!
}

还有我的解析器：

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？

【问题讨论】：

标签： graphql

【解决方案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

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

但你只有回报的人

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

您应该返回服务数组对象以使用 People 获取该返回对象

People: () => 返回类似下面的结果。

[{
    id: '5c5ac87e5e4d85ae77de9c50'
    Xid: 'xx'
    firstName: 'yyy'
    lastName: 'zzz'
    email: 'xxx@gmail.com'
    serviceId: '5c5ac87e5e4d85ae77de9c51'
    apps: ['testing', 'google']
    service: [{
        id: 5c5ac78d5e4d85ae77de9c4f,
        name: 'Tester',
        location: 'US America',
    }, {}]
}, {}]

如果您使用 mongoose，请使用 aggregation 从 mongoBD 获取数据。

【讨论】：

感谢您的回答。你能帮我实现聚合吗，我不明白。
非常抱歉，您所说的“人和服务的模态”是什么意思？
people.js: const mongoose = require('mongoose'); const Schema = mongoose.Schema; const peopleSchema = new Schema({ Xid: { type: String }, firstName: { type: String }, lastName: { type: String }, email: { type: String }, apps: { type: Array }, serviceId: { type: String } },{ versionKey: false }) module.exports = mongoose.model('people', peopleSchema);
service.js const mongoose = require('mongoose'); const Schema = mongoose.Schema; const serviceSchema = new Schema({ name: { type: String }, location: { type: String } },{ versionKey: false }) module.exports = mongoose.model('service', serviceSchema);
对于聚合或填充，您必须参考建立关系 serviceId: { type: Schema.Types.ObjectId, ref: 'service' },

【解决方案2】：

您需要在服务中传递与您的猫鼬模型相同的位置对象的引用（我假设您使用的是猫鼬）。传递引用后，您可以在服务查询Refer this中添加 populate(refernce).exec()

【讨论】：

谢谢，但作为一个菜鸟，我会问：你能发布一个代码示例吗？