为每个键添加前缀:

const obj = {foo: 'bar'}

const altObj = Object.fromEntries(
  Object.entries(obj).map(([key, value]) => 
    // Modify key here
    [`x-${key}`, value]
  )
)

// altObj = {'x-foo': 'bar'}

虽然这并不是一个更好的重命名键的解决方案，但它提供了一种快速简单的ES6方法来重命名对象中的所有键，同时不改变它们所包含的数据。

let b = {a: ["1"], b:["2"]};
Object.keys(b).map(id => {
  b[`root_${id}`] = [...b[id]];
  delete b[id];
});
console.log(b);

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

我的方法，改编好的@Mulhoon typescript帖子，用于更改多个键:

const renameKeys = <
    TOldKey extends keyof T,
    TNewkey extends string,
    T extends Record<string, unknown>
>(
  keys:  {[ key: string]: TNewkey extends TOldKey ? never : TNewkey },
  obj: T
) => Object
    .keys(obj)
    .reduce((acc, key) => ({
        ...acc,
        ...{ [keys[key] || key]: obj[key] }
    }), {});

renameKeys({id: 'value', name: 'label'}, {id: 'toto_id', name: 'toto', age: 35});

如果你想保留迭代顺序(插入的顺序)，这里有一个建议:

const renameObjectKey = (object, oldName, newName) => {

  const updatedObject = {}

  for(let key in object) {
      if (key === oldName) {
          newObject[newName] = object[key]
      } else {
          newObject[key] = object[key]
      }
  }

  object = updatedObject
}

如果你要改变源对象，ES6可以在一行中完成。

delete Object.assign(o, {[newKey]: o[oldKey] })[oldKey];

如果你想创建一个新对象，可以用两行。

const newObject = {};
delete Object.assign(newObject, o, {[newKey]: o[oldKey] })[oldKey];