Question

我有这个例子，我怎么能让它显示正确的答案。如果每个人都在线，则全局状态为在线状态。如果有人离开，全球状态就会消失。如果每个人都处于离线状态 - ＆gt;脱机。

var group, part1, part2, part3;

group = 'Group Status';
part1 = 'online';
part2 = 'away';
part3 = 'offline';

if( part1 === 'online' || part2 === 'online' || part3 === 'online' ){
    group = 'Online';
    console.log('All users are online. Group status: ' + group);
}else if( part1 === 'away' || part2 === 'away' || part3 === 'away' ){
    group = 'Away';
    console.log('One of the users is away. Group status: ' + group);
}else if( part1 === 'offline' || part2 === 'offline' || part3 === 'offline' ){
    group = 'Offline';
    console.log('One of the users is offline. Group status: ' + group);
}else{
    group = 'Not found';
    console.log('Status not found. Group status: ' + group);
}

答案：此代码的优先级是在线＆gt;离开＆gt;脱机。

我替换了它，它按我的意愿工作：离线＆gt;离开＆gt;在线。

var group, part1, part2, part3;
group = 'Group Status';
part1 = 'online';
part2 = 'away';
part3 = 'away';


if( part1 === 'offline' || part2 === 'offline' || part3 === 'offline' ){
    group = 'Offline';
    console.log('One of the users is offline. Group status: ' + group);
}else if( part1 === 'away' || part2 === 'away' || part3 === 'away' ){
    group = 'Away';
    console.log('One of the users is away. Group status: ' + group);
}else if( part1 === 'online' || part2 === 'online' || part3 === 'online' ){
    group = 'Online';
    console.log('All users are online. Group status: ' + group);
}else{
    group = 'Not found';
    console.log('Status not found. Group status: ' + group);
}

Answer 1

以下是我认为你想做的事：http://es6fiddle.net/itfgy2xt/

为组部件使用数组很有帮助，因此您可以使用数组方法，即Array.prototype.every和Array.prototype.indexOf。请注意，every是IE9 +。

仅当每个元素的函数返回every并且true返回数组中元素的索引或true时，

indexOf才返回-1找不到元素。

var groupStatus  = 'Group Status',
    parts = [];

parts[0] = 'online';
parts[1] = 'away';
parts[2] = 'offline';

// check if everyone is online
if (parts.every(function (el) { return el === 'online'; })) {
    groupStatus = 'Online';
}
// check if at least someone is online or away
else if (parts.indexOf('online') >= 0 || parts.indexOf('away') >= 0) {
    groupStatus = 'away';
}
else {
    groupStatus = 'offline';
}

console.log('Group status: ' + groupStatus);

Answer 2

您可以将所有部分移动到一个数组中，并使用Array#some和Array#every查看所需状态。

＆＃13;

var group = 'Group Status',
    parts = ['online', 'away', 'offline'];

if (parts.some(function (a) { return a === 'offline'; })) {
    group = 'Offline';
    console.log('One of the users is offline. Group status: ' + group);
} else if (parts.some(function (a) { return a === 'away'; })) {
    group = 'Away';
    console.log('One of the users is away. Group status: ' + group);
} else if (parts.every(function (a) { return a === 'online'; })) {
    group = 'Online';
    console.log('All users are online. Group status: ' + group);
} else {
    group = 'Not found';
    console.log('Status not found. Group status: ' + group);
}

＆＃13;

ES6

＆＃13;

var group = 'Group Status',
    parts = ['online', 'away', 'offline'],
    check = string => item => item === string;

if (parts.some(check('offline'))) {
    group = 'Offline';
    console.log('One of the users is offline. Group status: ' + group);
} else if (parts.some(check('away'))) {
    group = 'Away';
    console.log('One of the users is away. Group status: ' + group);
} else if (parts.every(check('online'))) {
    group = 'Online';
    console.log('All users are online. Group status: ' + group);
} else {
    group = 'Not found';
    console.log('Status not found. Group status: ' + group);
}

＆＃13;

如何得到正确的答案

2 个答案: