简单地这么做会有什么问题吗?

someObject = {...someObject, [newKey]: someObject.oldKey}
delete someObject.oldKey

如果愿意，可以将其包装在函数中:

const renameObjectKey = (object, oldKey, newKey) => {
    // if keys are the same, do nothing
    if (oldKey === newKey) return;
    // if old key doesn't exist, do nothing (alternatively, throw an error)
    if (!object.oldKey) return;
    // if new key already exists on object, do nothing (again - alternatively, throw an error)
    if (object.newKey !== undefined) return;

    object = { ...object, [newKey]: object[oldKey] };
    delete object[oldKey];

    return { ...object };
};

// in use
let myObject = {
    keyOne: 'abc',
    keyTwo: 123
};

// avoids mutating original
let renamed = renameObjectKey(myObject, 'keyTwo', 'renamedKey');

console.log(myObject, renamed);
// myObject
/* {
    "keyOne": "abc",
    "keyTwo": 123,
} */

// renamed
/* {
    "keyOne": "abc",
    "renamedKey": 123,
} */

这里的大多数答案都无法维持JS对象键值对的顺序。例如，如果您在屏幕上有一种希望修改的对象键-值对形式，那么保持对象条目的顺序就很重要。

ES6循环JS对象并将键值对替换为具有修改过的键名的新键值对的方法如下:

let newWordsObject = {};

Object.keys(oldObject).forEach(key => {
  if (key === oldKey) {
    let newPair = { [newKey]: oldObject[oldKey] };
    newWordsObject = { ...newWordsObject, ...newPair }
  } else {
    newWordsObject = { ...newWordsObject, [key]: oldObject[key] }
  }
});

该解决方案通过在旧条目的位置上添加新条目来保留条目的顺序。

就我个人而言，重命名对象中的键而不实现额外的沉重插件和轮子的最有效的方法:

var str = JSON.stringify(object);
str = str.replace(/oldKey/g, 'newKey');
str = str.replace(/oldKey2/g, 'newKey2');

object = JSON.parse(str);

如果对象具有无效的结构，还可以将其封装在try-catch中。工作完美无缺:)

一般来说，如果你想获得一个新的对象(不改变原来的对象)，根据keyMap重命名键-你可以使用以下基于lodash mapKeys的实现:

const {mapKeys} = require('lodash');
const renameKeys = (obj, keyMap) => _.mapKeys(obj, (value, key) => keyMap[key] || key);

使用的例子:

renameKeys({a: 1, b: 2, c: 3}, {c: 'p', a: 'm'})
> {m: 1, b: 2, p: 3}

我想这么做

const originalObj = {
  a: 1,
  b: 2,
  c: 3, // need replace this 'c' key into 'd'
};

const { c, ...rest } = originalObj;

const newObj = { ...rest, d: c };

console.log({ originalObj, newObj });

function iterate(instance) {
  for (let child of instance.tree_down) iterate(child);

  instance.children = instance.tree_down;
  delete instance.tree_down;
}

iterate(link_hierarchy);

console.log(link_hierarchy);