如何将JavaScript对象的键作为ID值添加到对象内部？

我有如下JavaScript对象：

{
    "xx2b": {
        "county": "Baringo",
        "town": "Kabarnet",
        "saccoRegistration": "BO2/08/009",
        "saccoName": "Baringo2"
    },
    "QQDa": {
        "saccoRegistration": "Ba/09/009",
        "town": "Mogotio",
        "county": "Baringo",
        "saccoName": "Baringo1"
    }
}

需要将每个子对象对应的外层键名，设置为该子对象内部的id属性值，预期结果如下：

{
    "xx2b": {
        "id": "xx2b",
        "county": "Baringo",
        "town": "Kabarnet",
        "saccoRegistration": "BO2/08/009",
        "saccoName": "Baringo2"
    },
    "QQDa": {
        "id":"QQDa",
        "saccoRegistration": "Ba/09/009",
        "town": "Mogotio",
        "county": "Baringo",
        "saccoName": "Baringo1"
    }
}

解决方案 #

方法一：使用Object.entries() + Array.reduce() #

这种写法属于函数式风格，代码简洁高效：

const originalObj = {
    "xx2b": {
        "county": "Baringo",
        "town": "Kabarnet",
        "saccoRegistration": "BO2/08/009",
        "saccoName": "Baringo2"
    },
    "QQDa": {
        "saccoRegistration": "Ba/09/009",
        "town": "Mogotio",
        "county": "Baringo",
        "saccoName": "Baringo1"
    }
};

const resultObj = Object.entries(originalObj).reduce((acc, [key, value]) => {
    // 复制原对象的属性并添加id
    acc[key] = { ...value, id: key };
    return acc;
}, {});

console.log(resultObj);

方法二：使用for...in循环 #

这种写法更直观，适合理解基础遍历逻辑：

const originalObj = {
    "xx2b": {
        "county": "Baringo",
        "town": "Kabarnet",
        "saccoRegistration": "BO2/08/009",
        "saccoName": "Baringo2"
    },
    "QQDa": {
        "saccoRegistration": "Ba/09/009",
        "town": "Mogotio",
        "county": "Baringo",
        "saccoName": "Baringo1"
    }
};

const resultObj = {};
for (const key in originalObj) {
    // 仅遍历对象自身的属性，避免原型链干扰
    if (originalObj.hasOwnProperty(key)) {
        resultObj[key] = { ...originalObj[key], id: key };
    }
}

console.log(resultObj);

说明 #

两种方法都通过**展开运算符(...)**复制原对象的属性，避免直接修改原对象（如果需要修改原对象，可以直接给originalObj[key].id = key，但通常推荐返回新对象保持原数据不变）。

内容的提问来源于stack exchange，提问作者steve gaikia