Question

我有嵌套对象用户

let users = {
    '0LsRDanmw1s1KBvZAAAC': 
    { 
        name: 'user1',
        room: 'Lobby' 
    },
    'Qvd9v0EStnwNe11mAAAD': 
    { 
        name: 'user2',
        room: 'Music' 
    },

};

我希望获得room === 'Lobby'

的用户数组

let onlineUsers = ['user1'];

提前致谢

编辑：很多好的解决方案谢谢你们！

Answer 1

获取值，执行.filter()然后.map()。

＆＃13;

let users = {    '0LsRDanmw1s1KBvZAAAC':     {         name: 'user1',        room: 'Lobby'     },    'Qvd9v0EStnwNe11mAAAD':     {         name: 'user2',        room: 'Music'     },};

var onlineUsers = Object.values(users)
                        .filter(u => u.room === 'Lobby')
                        .map(u => u.name);

console.log(onlineUsers );

＆＃13;

.as-console-wrapper { max-height: 100% !important; top: 0; }

＆＃13;

或者您可以获取值并执行.reduce()。

＆＃13;

let users = {    '0LsRDanmw1s1KBvZAAAC':     {         name: 'user1',        room: 'Lobby'     },    'Qvd9v0EStnwNe11mAAAD':     {         name: 'user2',        room: 'Music'     },};

var onlineUsers = Object.values(users)
                        .reduce((a, u) => (u.room === 'Lobby' ? [...a, u.name] : a), []);


console.log(onlineUsers);

＆＃13;

.as-console-wrapper {
  max-height: 100% !important;
  top: 0;
}

＆＃13;

Answer 2

您可以获取数组的值，根据需要进行过滤，并仅映射名称。

＆＃13;

var users = { '0LsRDanmw1s1KBvZAAAC': { name: 'user1', room: 'Lobby' }, 'Qvd9v0EStnwNe11mAAAD': { name: 'user2', room: 'Music' } },
    result = Object
        .values(users)
        .filter(({ room }) => room === 'Lobby')
        .map(({ name }) => name);
        
console.log(result);

＆＃13;

Answer 3

Object.values，filter和map的组合：

＆＃13;

let users = {'0LsRDanmw1s1KBvZAAAC': {   name: 'user1',room: 'Lobby' },'Qvd9v0EStnwNe11mAAAD':{ name: 'user2',room: 'Music' },};

const res = Object.values(users)
     .filter( o => o.room === 'Lobby')
     .map( o => o.name );
     
console.log(res);

＆＃13;

Answer 4

使用 ES8 Object.values()获取对象的所有值，然后应用.filter和.map

let users = {
    '0LsRDanmw1s1KBvZAAAC': 
    { 
        name: 'user1',
        room: 'Lobby' 
    },
    'Qvd9v0EStnwNe11mAAAD': 
    { 
        name: 'user2',
        room: 'Music' 
    },

};

var onlineUsers = Object.values(users).filter(el => el.room==="Lobby").map(el => el.name);

console.log(onlineUsers);

Answer 5

我仍然喜欢简单的for循环的旧学校，旧浏览器安全版本，所以为了完整起见：

var users = {
    '0LsRDanmw1s1KBvZAAAC': 
    { 
        name: 'user1',
        room: 'Lobby' 
    },
    'Qvd9v0EStnwNe11mAAAD': 
    { 
        name: 'user2',
        room: 'Music' 
    },
};

var lobbyUsers = [];
for (var prop in users) {
    if (users.hasOwnProperty(prop)) {
        if (users[prop].room === 'Lobby') {
            lobbyUsers.push(users[prop].name);
        }
    }
}
console.log(lobbyUsers);

从嵌套对象中获取数组

5 个答案: