首页 > 解决方案 > 映射函数返回 if

问题描述

我试图只返回与下面代码中的 id 匹配的用户。仅当 id 不匹配时,如何使其返回“不工作”?

const database = {
    users: [
        {
            id: '123',
            name: 'Shevy',
            email: 'shevy@gmail.com',
            password: 'soccer',
            entries: 0,
            joined: new Date()
        },

        {
            id: '124',
            name: 'Benny',
            email: 'benny@gmail.com',
            password: 'java',
            entries: 0,
            joined: new Date()
        }
    ]
}

let prime = 123

const mapArray = database.users.map((users) => {
      if (Number(users.id) === prime) {
          return users;
      }else {
          return "Not Working"
      }
             
    });
console.log(mapArray); 

这返回

[ { id: '123', 
    name: 'Shevy', 
    email: 'shevy@gmail.com', 
    password: 'soccer', 
    entries: 0, 
    joined: Thu Aug 13 2020 20:00:14 GMT+0100 (West Africa Standard Time) }, 
  'Not Working' ] 

我只想返回匹配的用户。

标签: javascriptarrays

解决方案


如果没有元素匹配,您可以使用Array#findnullish合并运算符来提供默认值。

const database = {
    users: [
        {
            id: '123',
            name: 'Shevy',
            email: 'shevy@gmail.com',
            password: 'soccer',
            entries: 0,
            joined: new Date()
        },
        {
            id: '124',
            name: 'Benny',
            email: 'benny@gmail.com',
            password: 'java',
            entries: 0,
            joined: new Date()
        }
    ]
};
const getUser = search => database.users.find(({id})=>+id===search) ?? 'Not working';
console.log(getUser(123));
console.log(getUser(9999));


推荐阅读