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/

从对象数组

function getSum(array, column)
  let values = array.map((item) => parseInt(item[column]) || 0)
  return values.reduce((a, b) => a + b)
}

foo = [
  { a: 1, b: "" },
  { a: null, b: 2 },
  { a: 1, b: 2 },
  { a: 1, b: 2 },
]

getSum(foo, a) == 3
getSum(foo, b) == 6

它在TypeScript和JavaScript中为我工作:

令LST = [ {说明:'高级'，价格:10}， {说明:“成人”，价格:20}， {描述:“儿童”，价格:30} ]; 令sum = lst。Map (o => o.price)。Reduce ((a, c) => {return a + c}); console.log(总和);

希望对大家有用。

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.

提高可读性和使用Map和Reduce的替代方案:

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

const sum = traveler
  .map(item => item.amount)
  .reduce((prev, curr) => prev + curr, 0);

可重用功能:

const calculateSum = (obj, field) => obj
  .map(items => items.attributes[field])
  .reduce((prev, curr) => prev + curr, 0);

我想我应该在这一点上发表我的意见:这是那些应该始终是纯函数式的操作之一，不依赖于任何外部变量。有几个已经给出了很好的答案，使用减法是这里要走的路。

既然我们大多数人已经可以使用ES2015语法，下面是我的建议:

const sumValues = (obj) => Object.keys(obj).reduce((acc, value) => acc + obj[value], 0);

我们把它变成了一个不可变函数。reduce在这里所做的很简单: 从累加器的值0开始，并将当前循环项的值加到累加器中。

函数式编程和ES2015太棒了!：）