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

Question

我正在开发一个使用GraphQL查询OrientDB图数据库的web应用程序。它使用Apollo服务器来解决传入的GraphQL查询。

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

// 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;

和关联的冲突解决程序：

//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，当我尝试实现上面的类型定义和解析器时，我收到了一个字符串列表，这些字符串都是“[object object]”，而不是实际的字符串：

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

我尝试向解析器添加一些代码，这些代码将遍历每个对象并创建要返回的正确字符串列表：

// 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');
        }
    }
};

但这并不起作用，而是返回了一个null值：

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

坦率地说，我很困惑为什么我不能通过这个解析器获得一个简单的字符串列表来填充。也许我遗漏了一些明显的东西--任何洞察力都是非常感激的！

Answer 1

您最初的方法没有像预期的那样工作，因为您返回的是一个对象数组。第二次尝试返回null，因为您没有返回解析器中的任何内容。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，所以您只需返回该值就足够了。但是我们可以使用map使这段代码更容易阅读，如下所示：

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)
}