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,
} */

如果你想保持对象的相同顺序

changeObjectKeyName(objectToChange, oldKeyName: string, newKeyName: string){
  const otherKeys = cloneDeep(objectToChange);
  delete otherKeys[oldKeyName];

  const changedKey = objectToChange[oldKeyName];
  return  {...{[newKeyName] : changedKey} , ...otherKeys};

}

使用方法:

changeObjectKeyName ( {'a' : 1}, 'a', 'A');

如果你不想改变你的数据，考虑这个函数…

renameProp = (oldProp, newProp, { [oldProp]: old, ...others }) => ({
  [newProp]: old,
  ...others
})

Yazeed Bzadough的详细解释 https://medium.com/front-end-hacking/immutably-rename-object-keys-in-javascript-5f6353c7b6dd

下面是一个typescript友好的版本:

// These generics are inferred, do not pass them in.
export const renameKey = <
  OldKey extends keyof T,
  NewKey extends string,
  T extends Record<string, unknown>
>(
  oldKey: OldKey,
  newKey: NewKey extends keyof T ? never : NewKey,
  userObject: T
): Record<NewKey, T[OldKey]> & Omit<T, OldKey> => {
  const { [oldKey]: value, ...common } = userObject

  return {
    ...common,
    ...({ [newKey]: value } as Record<NewKey, T[OldKey]>)
  }
}

它将防止您破坏现有的键或将其重命名为相同的东西

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。

使用对象解构和展开运算符的变体:

const old_obj = {
    k1: `111`,
    k2: `222`,
    k3: `333`
};    

// destructuring, with renaming. The variable 'rest' will hold those values not assigned to kA, kB, or kC.
const {
    k1: kA, 
    k2: kB, 
    k3: kC,
    ...rest
} = old_obj;
    

// now create a new object, with the renamed properties kA, kB, kC; 
// spread the remaining original properties in the 'rest' variable
const newObj = {kA, kB, kC, ...rest};

对于一个键，这可以很简单:

const { k1: kA, ...rest } = old_obj;
const new_obj = { kA, ...rest }

你也可能喜欢更“传统”的风格:

const { k1, ...rest } = old_obj
const new_obj = { kA: k1, ...rest}

const clone = (obj) => Object.assign({}, obj);

const renameKey = (object, key, newKey) => {

    const clonedObj = clone(object);
  
    const targetKey = clonedObj[key];
  
  
  
    delete clonedObj[key];
  
    clonedObj[newKey] = targetKey;
  
    return clonedObj;
     };

  let contact = {radiant: 11, dire: 22};





contact = renameKey(contact, 'radiant', 'aplha');

contact = renameKey(contact, 'dire', 'omega');



console.log(contact); // { aplha: 11, omega: 22 };