JavaScript按id合并数组重复对象 保留字段结构适配payroll系统CSV导出

实现思路 #

• 用Map结构做id去重的映射表，天然保留插入顺序，保证最终数组顺序和原数组首次出现id的顺序一致
• 合并时仅当已有对象的对应字段值为空时，才用新对象的字段值覆盖，避免空值抹除已有有效数据
• 直接基于首个出现的同id对象修改，完全保留原始字段顺序，适配后续CSV导出要求

实现代码 #

function mergeDuplicateEmployees(arr) {
  const idMap = new Map();
  for (const item of arr) {
    if (!idMap.has(item.id)) {
      // 首次出现的id，完整存入保留所有字段和顺序
      idMap.set(item.id, { ...item });
      continue;
    }
    const existing = idMap.get(item.id);
    // 遍历当前对象所有字段，仅空值才覆盖
    for (const key in item) {
      if (existing[key] === '' || existing[key] == null) {
        existing[key] = item[key];
      }
    }
  }
  // 转数组保留插入顺序
  return Array.from(idMap.values());
}

// 调用示例
const originalArr = [
  { id: 'MARS03', name: 'Employee 3', T1: 9.23, A6: '', SAT: '', SUN: '', PHW: '', SAN: '', COMM:'', BON: '', other: '', quantity: '', leaveDays: '', from: '', to: '', reason: '', costCentre: '', message: '' },
  { id: 'THOM01', name: 'Employee 4', T1: '', A6: '', SAT: 4.12, SUN: '', PHW: '', SAN: '', COMM: '', BON: '', other: '', quantity: '', leaveDays: '', from: '', to: '', reason: '', costCentre: '', message: '' },
  { id: 'THOM01', name: 'Employee 4', T1: '', A6: '', SAT: '', SUN: 6.12, PHW: '', SAN: '', COMM: '', BON: '', other: '', quantity: '', leaveDays: '', from: '', to: '', reason: '', costCentre: '', message: '' },
  { id: 'THOM01', name: 'Employee 4', T1: 15, A6: '', SAT: '', SUN: '', PHW: '', SAN: '', COMM: '', BON: '', other: '', quantity: '', leaveDays: '', from: '', to: '', reason: '', costCentre: '', message: '' }
];
const mergedArr = mergeDuplicateEmployees(originalArr);
console.log(mergedArr);

代码说明 #

• 如果你的字段空值定义包含undefined、null等情况，可以调整判断条件适配，比如if (!existing[key] && existing[key] !== 0)避免把数值0误判为空值
• 首次存入Map时浅拷贝原始对象，避免修改原数组数据，如果不需要保留原数组可以直接存item本身

内容的提问来源于stack exchange，提问作者davepiqch