如果有一个公共密钥，如何对json数组键的值求和

Question

我有一个像这样的json对象。 每个json对象都可以有共同的Product名称，例如orgs[0]＆amp; orgs[3]具有相同的产品名称。

"orgs":
    [{
    "Budget Actual Consumption": "12.00",
    "Budget Planned Consumption": "50.00",
    "Product": "Loyalty CO-brand"

}, {
    "Budget Actual Consumption": "11.00",
    "Budget Planned Consumption": "60.00",
    "Product": "Loaylty Rebates"
}, {
    "Budget Actual Consumption": "10.00",
    "Budget Planned Consumption": "7.00",
    "Product": "Loaylty Rebates"
}, {

    "Budget Actual Consumption": "9.00",
    "Budget Planned Consumption": "8.00",
    "Product": "Loyalty CO-brand"
}]

是否可以创建一个新阵列，使其具有唯一的多个对象，每个对象都有唯一的产品名称＆amp;其他关键将总结

例如

var someArray = [{
   name:"Loyalty CO-brand",
   bac:"21" //12+9
   pac:"58" //50+8
},{
  name:"Loaylty Rebates",
   bac:"21" //10+11
   pac:"67" //60+7 

}]

我首先尝试创建一个唯一的Product数组，然后在另一个内部使用forEach两次。但我无法成功完成，因为我不确定如果匹配Products

，我将如何添加其他键值

Answer 1

假设结果应按Product分组，您可以使用对象作为结果数组对象的哈希表。

如果省略获取字符串而不是数字的要求，则算法会更短。

var data = { "orgs": [{ "Budget Actual Consumption": "12.00 ", "Budget Planned Consumption": "50.00 ", "Product": "Loyalty CO-brand" }, { "Budget Actual Consumption": "11.00 ", "Budget Planned Consumption": "60.00 ", "Product": "Loaylty Rebates" }, { "Budget Actual Consumption": "10.00 ", "Budget Planned Consumption": "7.00 ", "Product": "Loaylty Rebates" }, { "Budget Actual Consumption": "9.00 ", "Budget Planned Consumption": "8.00 ", "Product": "Loyalty CO-brand" }] },
    grouped = [];

data.orgs.forEach(function (a) {
    if (!this[a.Product]) {
        this[a.Product] = { name: a.Product, bac: '0', pac: '0' };
        grouped.push(this[a.Product]);
    }
    this[a.Product].bac = (+this[a.Product].bac + +a['Budget Actual Consumption']).toString();
    this[a.Product].pac = (+this[a.Product].bac + +a['Budget Planned Consumption']).toString();

}, Object.create(null));

console.log(grouped);

Answer 2

使用var org = { "orgs": [{ "Budget Actual Consumption": "12.00 ", "Budget Planned Consumption": "50.00 ", "Product": "Loyalty CO-brand" }, { "Budget Actual Consumption": "11.00 ", "Budget Planned Consumption": "60.00 ", "Product": "Loaylty Rebates" }, { "Budget Actual Consumption": "10.00 ", "Budget Planned Consumption": "7.00 ", "Product": "Loaylty Rebates" }, { "Budget Actual Consumption": "9.00 ", "Budget Planned Consumption": "8.00 ", "Product": "Loyalty CO-brand" }] }, newArr = [], bac = 'Budget Actual Consumption', pac = 'Budget Planned Consumption'; org.orgs.forEach(function (o) { if (!this[o.Product]) { this[o.Product] = {name: o.Product, bac: +o[bac], pac: +o[pac]}; newArr.push(this[o.Product]); } else { this[o.Product]['bac'] += +o[bac]; this[o.Product]['pac'] += +o[pac]; } }, {}); console.log(JSON.stringify(newArr, 0, 4)); 函数的解决方案：

[
    {
        "name": "Loyalty CO-brand",
        "bac": 21,
        "pac": 58
    },
    {
        "name": "Loaylty Rebates",
        "bac": 21,
        "pac": 67
    }
]

输出：

{{1}}

Answer 3

我找到了Underscore groupBy

的一个很好的解决方案

http://codepen.io/therealplato/pen/KMWPPN

//foo = {"orgs":[...]}
console.log(foo);
grouped = _.groupBy(foo.orgs, grouper);
console.log(grouped);
function grouper(item) {
  return item["Product"];
}