如何在JavaScript/TypeScript中移除对象与另一对象重复的属性？

现有两个对象：
第一个基础对象：

{
  "id": "33343232",
  "createdAt": "2022-07-26T13:44:01.080Z",
  "updatedAt": "2022-07-26T13:45:31.000Z",
  "name": "Name Here",
  "description": "text"
}

第二个包含额外属性的对象：

"specificFeatures": {
    "id": "33343232",
    "createdAt": "2022-07-26T13:44:01.087Z",
    "updatedAt": "2022-07-26T13:45:31.000Z",
    "name": "Name Here",
    "description": "text",
    "coverage": "international",
    "income": 0,
    "observationIncome": ""
}

需要删除specificFeatures对象中与第一个对象属性键相同的属性，最终得到：

"specificFeatures": {
    "coverage": "international",
    "income": 0,
    "observationIncome": ""
}

以下是几种JavaScript的实现方式：

方法1：直接删除原对象的重复属性 #

先拿到第一个对象的所有属性键，遍历这些键从specificFeatures里删掉对应属性：

const baseObj = {
  "id": "33343232",
  "createdAt": "2022-07-26T13:44:01.080Z",
  "updatedAt": "2022-07-26T13:45:31.000Z",
  "name": "Name Here",
  "description": "text"
};

const targetObj = {
    "id": "33343232",
    "createdAt": "2022-07-26T13:44:01.087Z",
    "updatedAt": "2022-07-26T13:45:31.000Z",
    "name": "Name Here",
    "description": "text",
    "coverage": "international",
    "income": 0,
    "observationIncome": ""
};

// 获取基础对象的属性键集合
const baseKeys = Object.keys(baseObj);
// 遍历删除重复属性
baseKeys.forEach(key => {
  delete targetObj[key];
});

console.log(targetObj);
// 输出结果即为只保留差异属性的对象

方法2：创建新对象（不修改原对象） #

如果不想改动原specificFeatures对象，可生成一个新对象只保留不重复的属性：

const baseObj = {
  "id": "33343232",
  "createdAt": "2022-07-26T13:44:01.080Z",
  "updatedAt": "2022-07-26T13:45:31.000Z",
  "name": "Name Here",
  "description": "text"
};

const targetObj = {
    "id": "33343232",
    "createdAt": "2022-07-26T13:44:01.087Z",
    "updatedAt": "2022-07-26T13:45:31.000Z",
    "name": "Name Here",
    "description": "text",
    "coverage": "international",
    "income": 0,
    "observationIncome": ""
};

const baseKeys = new Set(Object.keys(baseObj));
// 筛选出不在基础对象中的属性，构建新对象
const filteredObj = Object.fromEntries(
  Object.entries(targetObj).filter(([key]) => !baseKeys.has(key))
);

console.log(filteredObj);
// 输出目标结果

方法3：解构赋值快速过滤 #

若基础对象的属性固定且数量不多，用解构赋值+剩余参数能快速提取需要保留的属性：

const baseObj = {
  "id": "33343232",
  "createdAt": "2022-07-26T13:44:01.080Z",
  "updatedAt": "2022-07-26T13:45:31.000Z",
  "name": "Name Here",
  "description": "text"
};

const targetObj = {
    "id": "33343232",
    "createdAt": "2022-07-26T13:44:01.087Z",
    "updatedAt": "2022-07-26T13:45:31.000Z",
    "name": "Name Here",
    "description": "text",
    "coverage": "international",
    "income": 0,
    "observationIncome": ""
};

// 解构出基础对象的属性，剩余的就是要保留的
const { id, createdAt, updatedAt, name, description, ...filteredObj } = targetObj;

console.log(filteredObj);
// 得到目标结果

内容的提问来源于stack exchange，提问作者Renan Bessa