我正试图弄清楚如何在ES6中做到这一点......
我有这个对象数组..
const originalData=[
{"investor": "Sue", "value": 5, "investment": "stocks"},
{"investor": "Rob", "value": 15, "investment": "options"},
{"investor": "Sue", "value": 25, "investment": "savings"},
{"investor": "Rob", "value": 15, "investment": "savings"},
{"investor": "Sue", "value": 2, "investment": "stocks"},
{"investor": "Liz", "value": 85, "investment": "options"},
{"investor": "Liz", "value": 16, "investment": "options"}
];
..和这个新的对象数组,我想在其中添加每个人的投资类型(股票,期权,储蓄)的总价值。
const newData = [
{"investor":"Sue", "stocks": 0, "options": 0, "savings": 0},
{"investor":"Rob", "stocks": 0, "options": 0, "savings": 0},
{"investor":"Liz", "stocks": 0, "options": 0, "savings": 0}
];
我循环遍历originalData并在let ..
中保存“当前对象”的每个属性for (let obj of originalData) {
let currinvestor = obj.investor;
let currinvestment = obj.investment;
let currvalue = obj.value;
..but here I want to find the obect in newData that has the property = currinvestor (for the "investor" key)
...then add that investment type's (currinvestment) value (currvalue)
}
答案 0 :(得分:51)
newData.find(x => x.investor === investor)
整个代码:
const originalData = [
{ "investor": "Sue", "value": 5, "investment": "stocks" },
{ "investor": "Rob", "value": 15, "investment": "options" },
{ "investor": "Sue", "value": 25, "investment": "savings" },
{ "investor": "Rob", "value": 15, "investment": "savings" },
{ "investor": "Sue", "value": 2, "investment": "stocks" },
{ "investor": "Liz", "value": 85, "investment": "options" },
{ "investor": "Liz", "value": 16, "investment": "options" },
];
const newData = [
{ "investor": "Sue", "stocks": 0, "options": 0, "savings": 0 },
{ "investor": "Rob", "stocks": 0, "options": 0, "savings": 0 },
{ "investor": "Liz", "stocks": 0, "options": 0, "savings": 0 },
];
for (let {investor, value, investment} of originalData) {
newData.find(x => x.investor === investor)[investment] += value;
}
console.log(newData);

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答案 1 :(得分:0)
我会使用它的一些衍生形式:
var arrayFindObjectByProp = (arr, prop, val) => {
return arr.find( obj => obj[prop] == val );
};