Question

我正在开发一个Web应用程序，该应用程序使用GraphQL查询OrientDB图形数据库。它使用Apollo Server解析传入的GraphQL查询。

我想构建一个查询，该查询将简单地将每个“主题”对象的“名称”字段作为字符串列表返回。例如：

{
  "data": {
    "allTopicNames": [
      "Topic 1",
      "Topic 2",
      "Topic 3",
      "Topic 4"
    ]
  }
}

为此，我创建了一个类型定义：

// Imports: GraphQL
import { gql } from 'apollo-server-express';

// GraphQL: TypeDefs
const TYPEDEFS = gql`
type Query {
    allTopics: [Topic]
    topic(name: String): [Topic]
    allTopicNames: [String] //This is the new Type Definition -- we want a list of Strings
  }
type Topic {
    name: String
}
`;

// Exports
export default TYPEDEFS;

以及相关的解析器：

//Connect to OrientDB
var OrientJs = require('orientjs');

var server = OrientJs({
    host: "localhost",
    port: "2424",
    username: "root",
    password: "root"
});

var db = server.use({
    name: 'database',
    username: 'root',
    password: 'root'
});

// GraphQL: Resolvers
const RESOLVERS = {
    Query: {
        allTopics: () => {
            return db.query('SELECT FROM Topic ORDER BY name');
        },
        allTopicNames: () => {
            return db.query('SELECT name FROM Topic ORDER BY name'); //This is the new resolver
        },
        topic: (obj, args) => {
            return db.query('SELECT FROM Topic WHERE name=\'' + args.name + '\' LIMIT 1');
        }
    }
};

// Exports
export default RESOLVERS;

但是，当我尝试实现上述类型定义和解析器时，我收到的字符串列表都是“ [object Object]”，而不是实际的字符串：

{
  "data": {
    "allTopicNames": [
      "[object Object]",
      "[object Object]",
      "[object Object]",
      "[object Object]"
    ]
  }
}

我试图向解析器添加一些代码，该代码将遍历每个对象并创建适当的字符串列表以返回：

// GraphQL: Resolvers
const RESOLVERS = {
    Query: {
        allTopics: () => {
            return db.query('SELECT FROM Topic ORDER BY name');
        },
        allTopicNames: () => {
            let the_list_of_records = db.query('SELECT name FROM Topic ORDER BY name').then(res => { 
                let the_list_of_names = []; //We'll return a List of Strings using this
                for(var i = 0; i < res.length; i++){
                    the_list_of_names.push(res[i]['name']);
                }
                console.log(the_list_of_names);
                return the_list_of_names;
            });
        },
        topic: (obj, args) => {
            return db.query('SELECT FROM Topic WHERE name=\'' + args.name + '\' LIMIT 1');
        }
    }
};

但是这不起作用，导致返回空值：

{
  "data": {
    "allTopicNames": null
  }
}

我很困惑为什么我无法通过此解析器获得简单的字符串列表。也许我缺少明显的东西-任何见解都将不胜感激！

Answer 1

您的初始方法未按预期工作，因为您要返回一个对象数组。您的第二次尝试返回null，因为您在解析器内部未返回任何内容。您的解析器应始终返回一个值或将解析为该值的Promise，否则该字段的解析值将始终为null。

the_list_of_records的值将是一个Promise，因此您只需返回它就足够了。但是我们可以像这样使用map使此代码更易于阅读：

allTopicNames: () => {
  return db.query('SELECT name FROM Topic ORDER BY name').then(res => {
    return res.map(topic => topic.name)
  })
}

// using async/await
allTopicNames: async () => {
  await topics = await db.query('SELECT name FROM Topic ORDER BY name')
  return topics.map(topic => topic.name)
}

构建GraphQL解析器以返回字符串列表-接收[object Object]而不是字符串

1 个答案: