如何对比两个对象数组并生成包含差异的新对象数组？

需求说明 #

对比两个对象数组，生成包含新旧属性差异的新对象数组：当原数组有对应索引的对象时，保留新旧属性值；原数组无对应对象时，旧属性值设为null。

示例输入 #

const objA = [{"name":"Karthi","age":25,"education":"Masters","city":"Chennai","country":"India"}]
const objB = [{"name":"Meenakshi","age":28,"education":"UG","city":"Pune","country":"India"},{"name":"Mani","age":31,"education":"Masters","city":"Madurai","country":"India"}]

预期输出 #

[{"oldName":"Karthi","newName":"Meenakshi","oldAge":25,"newAge":28,"oldEducation":"Masters","newEducation":"UG","oldCity":"Chennai","newCity":"Pune","country":"India"},{"oldName":null,"newName":"Mani","oldAge":null,"newAge":31,"oldEducation":null,"newEducation":"Masters","oldCity":null,"newCity":"Madurai","country":"India"}]

代码实现 #

function compareObjectArrays(arrA, arrB) {
  return arrB.map((newObj, index) => {
    const result = {};
    const oldObj = arrA[index] || {};
    
    Object.keys(newObj).forEach(key => {
      // 按示例规则，country属性直接保留新值
      if (key === 'country') {
        result[key] = newObj[key];
        return;
      }
      // 生成新旧属性键名，旧值不存在时设为null
      const capitalizedKey = key.charAt(0).toUpperCase() + key.slice(1);
      result[`old${capitalizedKey}`] = oldObj[key] ?? null;
      result[`new${capitalizedKey}`] = newObj[key];
    });
    
    return result;
  });
}

// 测试示例
const objA = [{"name":"Karthi","age":25,"education":"Masters","city":"Chennai","country":"India"}];
const objB = [{"name":"Meenakshi","age":28,"education":"UG","city":"Pune","country":"India"},{"name":"Mani","age":31,"education":"Masters","city":"Madurai","country":"India"}];

console.log(compareObjectArrays(objA, objB));

逻辑说明 #

• 以目标数组arrB为基准，遍历每个对象生成差异项
• 匹配arrA对应索引的对象，无对应对象则用空对象兜底
• 对每个属性做差异化处理：
  • country属性直接保留新对象的值（符合示例规则）
  • 其他属性生成oldXxx和newXxx格式的键名，旧值不存在时设为null
• 最终返回整理后的差异对象数组

内容的提问来源于stack exchange，提问作者karthi