Graphql 查询总是返回 null

Posted 2023-03-08

【中文标题】Graphql 查询总是返回 null

【英文标题】：Graphql query is always returning null

【发布时间】：2021-01-15 12:21:21

【问题描述】：

我正在尝试使用 graphql 获取课程数据，但服务器总是返回 null 作为响应这是我在文件 server.js 中的代码：

var express=require('express');
const  graphqlHTTP  = require('express-graphql')
var buildSchema=require('graphql');

//graphql schema
var schema=buildSchema(`
type Query 
    course(id: Int!): Course
    courses(topic:String!):[Course]


type Course 
    id: Int
    title: String
    author: String
    description:String
    topic:String
    url: String

`);

var coursesData = [
    
        id: 1,
        title: 'The Complete Node.js Developer Course',
        author: 'Andrew Mead, Rob Percival',
        description: 'Learn Node.js by building real-world applications with Node, Express, MongoDB, Mocha, and more!',
        topic: 'Node.js',
        url: 'https://codingthesmartway.com/courses/nodejs/'
    ,
    
        id: 2,
        title: 'Node.js, Express & MongoDB Dev to Deployment',
        author: 'Brad Traversy',
        description: 'Learn by example building & deploying real-world Node.js applications from absolute scratch',
        topic: 'Node.js',
        url: 'https://codingthesmartway.com/courses/nodejs-express-mongodb/'
    ,
    
        id: 3,
        title: 'javascript: Understanding The Weird Parts',
        author: 'Anthony Alicea',
        description: 'An advanced JavaScript course for everyone! Scope, closures, prototypes, this, build your own framework, and more.',
        topic: 'JavaScript',
        url: 'https://codingthesmartway.com/courses/understand-javascript/'
    
]

//root resolver
var root= 
    course:getCourse,
    courses:getCourses
;


var getCourse= function (args)
    var id=args.id;
    console.log("delila")
    return coursesData.filter(course=>
        return course.id==id
    )[0]


var getCourses = function(args)
    if(args.topic)
        var topic=args.topic;
        return coursesData.filter(course=>
            course.topic===topic
        );

    
    else return coursesData
    

//create an experess server and graphql endpoint
var app=express();
app.use('/graphql', graphqlHTTP(
    schema: schema,
    rootValue:root,
    graphiql:true
));
app.listen(4000,()=>console.log("delila express graphql server running on localhost 4000"))

当我去localhost:4000/graphql获取我正在使用的数据时

query getSingleCourse($courseID: Int !)
  course(id:$courseID)
    title
    author
    description
    url
    topic
  



  "courseID": 3

但我不断得到结果null。看图

有人知道为什么会这样吗？服务器应该使用id 3 返回课程，但显然我缺少一些东西

【问题讨论】：

'args' 是第二个解析器 fn arg

【参考方案1】：

您应该先定义函数表达式，然后再使用它们。就是这个原因。

与函数声明不同，JavaScript 中的函数表达式不会被提升。在创建函数表达式之前不能使用它们：

见Function expression

例如

//...

var getCourse = function (args) 
  var id = args.id;
  console.log('delila');
  return coursesData.filter((course) => 
    return course.id == id;
  )[0];
;
var getCourses = function (args) 
  if (args.topic) 
    var topic = args.topic;
    return coursesData.filter((course) => course.topic === topic);
   else return coursesData;
;

//root resolver
var root = 
  course: getCourse,
  courses: getCourses,
;

//...