汇总对象内对象的单个属性的所有值答案

【问题标题】：Total all values of a single property from an object within an object汇总对象内对象的单个属性的所有值
【发布时间】：2016-08-14 00:27:22
【问题描述】：

function player(name, goals, games) {
    this.name = name;
    this.goals = goals;
    this.games = games;
}

var ricky = new player('Ricky', 7, 15);
var tom = new player('Tom', 15, 17);
var phil = new player('Phillip', 9, 14);
var jerry = new player('Jerry', 11, 15);
var randy = new player('Randy', 4, 16);
var sam = new player('Sam', 5, 11);


function locTeams(name, town, wins, playerOne, playerTwo, playerThree) {
    this.name = name;
    this.town = town;
    this.wins = wins;
    this.playerOne = playerOne;
    this.playerTwo = playerTwo;
    this.playerThree = playerThree;
}

var tigers = new locTeams('Great Tigers', 'Clayton', 9, ricky, tom);
var pantheon = new locTeams('The Pantheons', 'Brookedale', 8, jerry, randy);

teams = [tigers, pantheon];

var totalGoalsState =

我需要一种简单的方法让totalGoalsState 等于teams 数组中玩家的累计玩家目标。另外，我如何将playerThree 与其他新玩家之一（例如phil 或sam）填充到像老虎队或万神殿这样的球队之一。

【问题讨论】：

locTeams 的构造函数需要 5 个参数，你传递了 4 个。你也错过了this.playerThree = playerThree。
@AleksandarĐokić - 它并不真正需要 5 个参数，它可以有 5 个参数。
@adeneo 是的，但他没有在方法中的任何地方设置playerThree......他应该，因为它在构造函数中。
糟糕，忘了补充，我修好了。

标签： javascript arrays variables object new-operator

【解决方案1】：

这是将数组中所有团队的目标相加的一种方法，迭代和减少每个玩家、每个团队等的值。

function player(name, goals, games) {
    this.name = name;
    this.goals = goals;
    this.games = games;
}
function player(name, goals, games) {
    this.name = name;
    this.goals = goals;
    this.games = games;
}

var ricky = new player('Ricky', 7, 15);
var tom = new player('Tom', 15, 17);
var phil = new player('Phillip', 9, 14);
var jerry = new player('Jerry', 11, 15);
var randy = new player('Randy', 4, 16);
var sam = new player('Sam', 5, 11);


function locTeams(name, town, wins, playerOne, playerTwo, playerThree) {
    this.name = name;
    this.town = town;
    this.wins = wins;
    this.playerOne = playerOne;
    this.playerTwo = playerTwo;
    this.playerThree = playerThree;
}

var tigers = new locTeams('Great Tigers', 'Clayton', 9, ricky, tom);
var pantheon = new locTeams('The Pantheons', 'Brookedale', 8, jerry, randy);

var teams = [tigers, pantheon];

function getGoals(team) {
		return Object.keys(team).map(function(k) {
    	return k.indexOf('player') === 0 ? 
          team[k] && "goals" in team[k] ? team[k].goals : 0 : 0;
    }).reduce(function(a,b) { return a+b });
}

var totalGoalsState = teams.reduce(function(a,b) {return getGoals(a) + getGoals(b)});

console.log(totalGoalsState)

【讨论】：

【解决方案2】：

如果您需要第三个玩家，只需将其添加到其他两个玩家所在的位置即可。请注意，我正在消除重复项，因为每个玩家只能计入一次总目标。

function player(name, goals, games) {
    this.name = name;
    this.goals = goals;
    this.games = games;
}

var ricky = new player('Ricky', 7, 15);
var tom = new player('Tom', 15, 17);
var phil = new player('Phillip', 9, 14);
var jerry = new player('Jerry', 11, 15);
var randy = new player('Randy', 4, 16);
var sam = new player('Sam', 5, 11);


function locTeams(name, town, wins, playerOne, playerTwo, playerThree) {
    this.name = name;
    this.town = town;
    this.wins = wins;
    this.playerOne = playerOne;
    this.playerTwo = playerTwo;
}

var tigers = new locTeams('Great Tigers', 'Clayton', 9, ricky, tom);
var pantheon = new locTeams('The Pantheons', 'Clayton', 9, jerry, randy);

teams = [tigers, pantheon];

var totalGoalsState = teams.reduce((p, c) => p.concat([c.playerOne, c.playerTwo]), [])
                           .reduce((p, c) => p.includes(c) ? p : p.concat(c), [])
                           .reduce((p, c) => p + c.goals, 0);

console.log(totalGoalsState);

【讨论】：