如何去除对象中zone字符串属性的重复值

可以通过以下步骤实现zone属性的动态去重：

1. 将zone字符串按, 分割为数组
2. 利用Set数据结构自动过滤重复值（Set不允许存储重复元素）
3. 将去重后的Set转回数组，再用, 拼接成目标格式的字符串
4. 把处理后的字符串赋值回原对象的zone属性

示例代码 #

// 原始对象
const originalObj = {
    "totalMinutes": "00:01",
    "exception": "Yes",
    "observation": [],
    "zone": "Zone 4, Zone 1, Zone 1, Zone 7, Zone 1, Zone 8, Zone 20, Zone 1, Zone 1",
    "startTime": "2023-01-03T06:22:17.298Z",
    "appVersion": "5.11.11",
    "month": "January",
    "userName": "Geeta Ammanola",
    "endLatlong": "0,0",
    "activity": "Rapid Tour",
    "profile": "Field Supervisor",
    "date": "20230103",
    "endTime": "2023-01-03T06:23:49.526Z",
    "status": "ABORTED",
    "rapidTourId": "bOiOmARnl7I8lGw1UAxN",
    "actualBOGHour": "00:01",
    "email": "geeta.ammanola.ext1@holcim.com",
    "intermediateBarcodeScanned": [],
    "region": "Asia",
    "opco": "Holcim Australia",
    "plant": "5010 Milton Concrete",
    "role": "country_admin",
    "bogMinutes": 20
};

// 处理zone去重
originalObj.zone = [...new Set(originalObj.zone.split(', '))].join(', ');

// 输出处理后的zone结果
console.log({ zone: originalObj.zone });

运行结果 #

{"zone": "Zone 4, Zone 1, Zone 7, Zone 8, Zone 20"}

这种方法无需硬编码匹配特定内容，完全适配动态变化的zone数据，完美解决replace()方法无法处理动态重复值的问题。

内容的提问来源于stack exchange，提问作者Ragesh Pikalmunde