合并对象数组，修改一些结果条目答案

【问题标题】：Merge Arrays of Objects, Modifying Some Resulting Entries合并对象数组，修改一些结果条目
【发布时间】：2019-06-09 04:36:00
【问题描述】：

我有以下代码，它在 JavaScript 中获取 2 个对象数组并比较它们并合并内容，返回第一个数组中的所有对象，以及第二个数组中的所有对象，其 id 不在第一个数组中。

但是对于来自第二个数组的那些项目，我想修改它在新数组中的条目以仅包含某些属性（例如，id & make）并添加一个新的占位符属性（例如，输入：“汽车”）。我相信这会使用地图功能，但我似乎无法让它工作。

    var cars1 = [
        {id: 1, make: "Ford",  model: "F150",  year: 2002},
        {id: 3, make: "Chevy", model: "Tahoe", year: 2003},
    ];

    var cars2 = [
        {id: 2, make: "Kia",    model: "Optima",  year: 2001},
        {id: 4, make: "Nissan", model: "Sentra",  year: 1982},
    ];

    const cars1IDs = new Set(cars1.map(({ id }) => id));
    const combined = [
      ...cars1,
      ...cars2.filter(({ id }) => !cars1IDs.has(id))
              .map(cars2 => cars2.id)
    ];
    combined.sort(({ id: aId }, {id: bId }) => aId - bId);
    console.log(combined);

我希望组合数组以以下内容结束：

var combined = [
    {id: 1, make: "Ford",   model: "F150",  year: 2002},
    {id: 2, make: "Kia",    type: "car"},
    {id: 3, make: "Chevy",  model: "Tahoe", year: 2003},
    {id: 4, make: "Nissan", type: "car"},
];

但结果却是：

var combined = [
    {id: 1, make: "Ford",   model: "F150",  year: 2002},
    {id: 3, make: "Chevy", model: "Tahoe", year: 2003},
    2,
    4
];

【问题讨论】：

标签： javascript dictionary filter

【解决方案1】：

在.map 中，您应该提取id 和make 属性，并返回一个仅包含这些属性以及type 的car 的对象：

var cars1 = [
    {id: 1, make: "Ford",  model: "F150",  year: 2002},
    {id: 3, make: "Chevy", model: "Tahoe", year: 2003},
];

var cars2 = [
    {id: 2, make: "Kia",    model: "Optima",  year: 2001},
    {id: 4, make: "Nissan", model: "Sentra",  year: 1982},
];

const cars1IDs = new Set(cars1.map(({ id }) => id));
const combined = [
  ...cars1,
  ...cars2.filter(({ id }) => !cars1IDs.has(id))
          .map(({ id, make }) => ({ id, make, type: 'car' }))
];
combined.sort((a, b) => a.id - b.id);
console.log(combined);

【讨论】：

这是完美的。谢谢！

【解决方案2】：

你应该改变地图功能这是代码。只有car2的地图功能是变化的

var cars1 = [
        { id: 1, make: "Ford", model: "F150", year: 2002 },
        { id: 3, make: "Chevy", model: "Tahoe", year: 2003 }
      ];

      var cars2 = [
        { id: 2, make: "Kia", model: "Optima", year: 2001 },
        { id: 4, make: "Nissan", model: "Sentra", year: 1982 }
      ];

      const cars1IDs = new Set(cars1.map(({ id }) => id));
      const combined = [
        ...cars1,
        ...cars2
          .filter(({ id }) => !cars1IDs.has(id))
          .map(car2 => {
              //add new prop type to car
              const tempCar2 = Object.assign(car2,{type:"car"});
              //remove year and model
              delete tempCar2.year                    
              delete tempCar2.model
              return tempCar2;
            //Object.assign(car2, { id: car2.id });
          })
      ];
      combined.sort(({ id: aId }, { id: bId }) => aId - bId);
      console.log(combined);

【讨论】：