你可以使用Array.prototype.reduce:

const sum = traveler.reduce((acc , val)=>{
   return acc + val.amount;
} ,0);

更新后的答案

由于向Array原型添加函数的所有缺点，我正在更新这个答案，以提供一种替代方案，使语法与问题中最初请求的语法相似。

class TravellerCollection extends Array {
    sum(key) {
        return this.reduce((a, b) => a + (b[key] || 0), 0);
    }
}
const traveler = new TravellerCollection(...[
    {  description: 'Senior', Amount: 50},
    {  description: 'Senior', Amount: 50},
    {  description: 'Adult', Amount: 75},
    {  description: 'Child', Amount: 35},
    {  description: 'Infant', Amount: 25 },
]);

console.log(traveler.sum('Amount')); //~> 235

原来的答案

因为它是一个数组，你可以添加一个函数到数组原型。

traveler = [
    {  description: 'Senior', Amount: 50},
    {  description: 'Senior', Amount: 50},
    {  description: 'Adult', Amount: 75},
    {  description: 'Child', Amount: 35},
    {  description: 'Infant', Amount: 25 },
];

Array.prototype.sum = function (prop) {
    var total = 0
    for ( var i = 0, _len = this.length; i < _len; i++ ) {
        total += this[i][prop]
    }
    return total
}

console.log(traveler.sum("Amount"))

小提琴:http://jsfiddle.net/9BAmj/

只是另一种说法，这就是原生JavaScript函数Map和Reduce的用途(Map和Reduce是许多语言中的强大功能)。

var traveler = [{description: 'Senior', Amount: 50},
                {description: 'Senior', Amount: 50},
                {description: 'Adult', Amount: 75},
                {description: 'Child', Amount: 35},
                {description: 'Infant', Amount: 25}];

function amount(item){
  return item.Amount;
}

function sum(prev, next){
  return prev + next;
}

traveler.map(amount).reduce(sum);
// => 235;

// or use arrow functions
traveler.map(item => item.Amount).reduce((prev, next) => prev + next);

注意:通过创建独立的小函数，我们可以再次使用它们。

// Example of reuse.
// Get only Amounts greater than 0;

// Also, while using Javascript, stick with camelCase.
// If you do decide to go against the standards, 
// then maintain your decision with all keys as in...

// { description: 'Senior', Amount: 50 }

// would be

// { Description: 'Senior', Amount: 50 };

var travelers = [{description: 'Senior', amount: 50},
                {description: 'Senior', amount: 50},
                {description: 'Adult', amount: 75},
                {description: 'Child', amount: 35},
                {description: 'Infant', amount: 0 }];

// Directly above Travelers array I changed "Amount" to "amount" to match standards.

function amount(item){
  return item.amount;
}

travelers.filter(amount);
// => [{description: 'Senior', amount: 50},
//     {description: 'Senior', amount: 50},
//     {description: 'Adult', amount: 75},
//     {description: 'Child', amount: 35}];
//     Does not include "Infant" as 0 is falsey.

使用reduce和解构来求和

const traveler = [
  { description: 'Senior', Amount: 50 },
  { description: 'Senior', Amount: 50 },
  { description: 'Adult', Amount: 75 },
  { description: 'Child', Amount: 35 },
  { description: 'Infant', Amount: 25 },
];

console.log(traveler.reduce((n, {Amount}) => n + Amount, 0))

我知道这个问题已经有了一个公认的答案，但我认为我应该加入一个使用数组的替代方案。Reduce，因为对数组求和是Reduce的规范示例:

$scope.sum = function(items, prop){
    return items.reduce( function(a, b){
        return a + b[prop];
    }, 0);
};

$scope.travelerTotal = $scope.sum($scope.traveler, 'Amount');

小提琴

下面是一个使用ES6箭头函数的一行程序。

const sumPropertyValue = (items, prop) => items.reduce((a, b) => a + b[prop], 0);

// usage:
const cart_items = [ {quantity: 3}, {quantity: 4}, {quantity: 2} ];
const cart_total = sumPropertyValue(cart_items, 'quantity');