如何将一个对象的值合并到对象数组中的一个元素？答案

【问题标题】：How can I merge value from one object to an element in an array of objects?如何将一个对象的值合并到对象数组中的一个元素？
【发布时间】：2022-11-22 13:06:15
【问题描述】：

我有两个数组——一个是字符串，一个是对象。我在第一个数组中找到重复项并进行计数。我想使用 Product2Id 作为键将我在计数对象中找到的整数值添加到 pbentry。老实说，我一直无法弄清楚，而且谷歌数小时也没有结果。

期望的结果：

[{"Id":"01u8D00000105oqQAA","Product2Id":"01t8D000001fDfjQAE","Count":"3"},
{"Id":"01u8D00000105oxQAA","Product2Id":"01t8D000001fDfqQAE","Count":"1"},
{"Id":"01u8D00000105p2QAA","Product2Id":"01t8D000001fDfvQAE","Count":"1"},
{"Id":"01u8D000003WBH5QAO","Product2Id":"01t1O000004XyR0QAK","Count":"2"},
{"Id":"01u8D000003WBH0QAO","Product2Id":"01t8D000001hKF1QAM","Count":"1"}....];

let counts = {}; 
let array = ["01t8D0000014jiuQAA", "01t5Y000006VydJQAS", "01t8D000001fDfjQAE", "01t8D000001fDfjQAE", "01t8D000001hKF1QAM", "01t1O000004XyR0QAK", "01t14000006956yAAA", "01t1O000004XyR0QAK", "01t8D000001fDfqQAE", "01t8D000001f1yeQAA", "01t8D000001fDfvQAE", "01t8D000001fDfjQAE"];
let pbentry = [{"Id":"01u8D000003WBHAQA4","Product2Id":"01t14000006956yAAA"},{"Id":"01u8D000003WBH5QAO","Product2Id":"01t1O000004XyR0QAK"}, {"Id":"01u8D000000zEfiQAE","Product2Id":"01t5Y000006VydJQAS"},{"Id":"01u8D000003WBGqQAO","Product2Id":"01t8D0000014jiuQAA"},
{"Id":"01u8D000003WBHyQAO","Product2Id":"01t8D000001f1yeQAA"},{"Id":"01u8D00000105oqQAA","Product2Id":"01t8D000001fDfjQAE"}, {"Id":"01u8D00000105oxQAA","Product2Id":"01t8D000001fDfqQAE"},{"Id":"01u8D00000105p2QAA","Product2Id":"01t8D000001fDfvQAE"}, {"Id":"01u8D000003WBH0QAO","Product2Id":"01t8D000001hKF1QAM"}];

array.forEach(function (x) { counts[x] = (counts[x] || 0) + 1; }); 

console.log(pbentry)

【问题讨论】：

标签： javascript

【解决方案1】：

下面介绍的是实现预期目标的一种可能方法。

代码片段

const myTransform = (countArr, infoArr) => (
  Object.values(        // extract values from intermediate result object below
    infoArr?.reduce(    // use ".reduce()" to generate a map/dictionary using 'Id' as key
      (acc, {Id, ...rest}) => {    // de-structure each item to obtain 'Id'
        acc[Id] ??= {Id, Count: 0, ...rest};    // nullish coalescing assignment
        // update "Count" by filtering 'countArr' and getting the number of times
        acc[Id]["Count"] = countArr?.filter(x => x === Id)?.length;
        // always return the "acc" (ie, accumulator of ".reduce()")
        return acc;
      },
      {}          // initialize "acc" as an empty object
    )
  )               // implicit return
);

let counts = {};
let myArray = ["01t8D0000014jiuQAA", "01t5Y000006VydJQAS", "01t8D000001fDfjQAE", "01t8D000001fDfjQAE", "01t8D000001hKF1QAM",
  "01t1O000004XyR0QAK", "01t14000006956yAAA", "01t1O000004XyR0QAK", "01t8D000001fDfqQAE", "01t8D000001f1yeQAA",
  "01t8D000001fDfvQAE", "01t8D000001fDfjQAE"
];


let pbentry = [{
    "Id": "01u8D000003WBHAQA4",
    "Product2Id": "01t14000006956yAAA"
  }, {
    "Id": "01u8D000003WBH5QAO",
    "Product2Id": "01t1O000004XyR0QAK"
  },
  {
    "Id": "01u8D000000zEfiQAE",
    "Product2Id": "01t5Y000006VydJQAS"
  }, {
    "Id": "01u8D000003WBGqQAO",
    "Product2Id": "01t8D0000014jiuQAA"
  },
  {
    "Id": "01u8D000003WBHyQAO",
    "Product2Id": "01t8D000001f1yeQAA"
  }, {
    "Id": "01u8D00000105oqQAA",
    "Product2Id": "01t8D000001fDfjQAE"
  },
  {
    "Id": "01u8D00000105oxQAA",
    "Product2Id": "01t8D000001fDfqQAE"
  }, {
    "Id": "01u8D00000105p2QAA",
    "Product2Id": "01t8D000001fDfvQAE"
  },
  {
    "Id": "01u8D000003WBH0QAO",
    "Product2Id": "01t8D000001hKF1QAM"
  }
];

console.log(
  'counting unique ids:
',
  myTransform(myArray, pbentry)
);

.as-console-wrapper { max-height: 100% !important; top: 0 }

解释

添加到上面的 sn-p 的内联 cmets。

【讨论】：

【解决方案2】：

这一行应该做到这一点，它只是获取您已经计算出的计数并将其附加到您的 pbentry 数组。：

pbentry = pbentry.map(pb => ({...pb, Count: counts[pb.Product2Id] || 0}))

在计算出计数后添加它：

let counts = {}; 
let array = ["01t8D0000014jiuQAA", "01t5Y000006VydJQAS", "01t8D000001fDfjQAE", "01t8D000001fDfjQAE", "01t8D000001hKF1QAM", "01t1O000004XyR0QAK", "01t14000006956yAAA", "01t1O000004XyR0QAK", "01t8D000001fDfqQAE", "01t8D000001f1yeQAA", "01t8D000001fDfvQAE", "01t8D000001fDfjQAE"];
let pbentry = [{"Id":"01u8D000003WBHAQA4","Product2Id":"01t14000006956yAAA"},{"Id":"01u8D000003WBH5QAO","Product2Id":"01t1O000004XyR0QAK"}, {"Id":"01u8D000000zEfiQAE","Product2Id":"01t5Y000006VydJQAS"},{"Id":"01u8D000003WBGqQAO","Product2Id":"01t8D0000014jiuQAA"},
{"Id":"01u8D000003WBHyQAO","Product2Id":"01t8D000001f1yeQAA"},{"Id":"01u8D00000105oqQAA","Product2Id":"01t8D000001fDfjQAE"}, {"Id":"01u8D00000105oxQAA","Product2Id":"01t8D000001fDfqQAE"},{"Id":"01u8D00000105p2QAA","Product2Id":"01t8D000001fDfvQAE"}, {"Id":"01u8D000003WBH0QAO","Product2Id":"01t8D000001hKF1QAM"}];

array.forEach(function (x) { counts[x] = (counts[x] || 0) + 1; }); 

pbentry = pbentry.map(pb => ({...pb, Count: counts[pb.Product2Id] || 0}))

console.log(pbentry)

【讨论】：