我正在努力合并 2 个对象。它们都有嵌套对象，任意键的值可以是空字符串。

const obj1 = { name: "Arthur", lastName: "King", address: {
  street: "",
  city: "Some city",
  zip: "26772",
}};

const obj2 = { name: "", lastName: "", address: {
  street: "Some street",
  city: "",
  zip: "",
}};

Object.assign 和 spread operator 不工作，因为它们正在进行浅合并，这不是我的情况，即使我单独合并嵌套对象也是如此。

问题在于，合并 2 个具有空字符串值的对象无法按预期工作，因为 Javascript 将空字符串视为不同于 undefined 或 null 的值。

const merged = {...obj1, ...obj2, address: {...obj1.address, ...obj2.address}}

The result will be

{
    "name": "",
    "lastName": "",
    "address": {
        "street": "Some street",
        "city": "",
        "zip": ""
    }
}

期望的结果可能是下面的结果，但不知何故它不会忽略空值并将它们视为值。

{
    "name": "",
    "lastName": "",
    "address": {
        "street": "Some street",
        "city": "Some city",
        "zip": "26772"
    }
}

解决方法之一是从对象中删除所有空字符串值，然后进行合并，但恕我直言，这完全是矫枉过正。

最佳答案

您可以通过获取条目并检查值是否为对象来创建一个新对象，然后将合并的嵌套对象作为值。

此方法仅适用于相同的对象结构。

const
    merge = (a, b) => Object.fromEntries(Object
        .entries(a)
        .map(([k, v]) =>  [k, v && typeof v === 'object'
            ? merge(v, b[k])
            : v || b[k]
        ])
    ),
    obj1 = { name: "Arthur", lastName: "King", address: { street: "", city: "Some city", zip: "26772" } },
    obj2 = { name: "", lastName: "", address: { street: "Some street", city: "", zip: "" } },
    merged = merge(obj1, obj2);

console.log(merged);

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https://stackoverflow.com/questions/71980908/