是否有一个聪明的(即优化)方法重命名一个关键在javascript对象?

一种非优化的方式是:

o[ new_key ] = o[ old_key ];
delete o[ old_key ];

当前回答

如果有人需要重命名属性列表:

function renameKeys(obj, newKeys) {
  const keyValues = Object.keys(obj).map(key => {
    const newKey = newKeys[key] || key;
    return { [newKey]: obj[key] };
  });
  return Object.assign({}, ...keyValues);
}

用法:

const obj = { a: "1", b: "2" };
const newKeys = { a: "A", c: "C" };
const renamedObj = renameKeys(obj, newKeys);
console.log(renamedObj);
// {A:"1", b:"2"}

其他回答

还有一种最强大的REDUCE方法。

数据= {\ key1:“value1”,key2:“value2”,key3:“value3”}; 键文件夹= {\ key1:“firstkey”,key2:“secondkey”,key3:“thirdkey” mappedData = Object.keys . .还原((obj,k) =>对象。assign(obj, {\ [keyMap]]:数据[k]},{\}; 控制台日志(mappedData); 第一个“value1”、“secondkey”、“value2”、“thirdkey”、“value3”……

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 };

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});

我只想用ES6(ES2015)的方式!

我们需要跟上时代!

const old_obj = { k1: `111`, k2: `222`, k3: `333` }; console.log(`old_obj =\n`, old_obj); // {k1: "111", k2: "222", k3: "333"} /** * @author xgqfrms * @description ES6 ...spread & Destructuring Assignment */ const { k1: kA, k2: kB, k3: kC, } = {...old_obj} console.log(`kA = ${kA},`, `kB = ${kB},`, `kC = ${kC}\n`); // kA = 111, kB = 222, kC = 333 const new_obj = Object.assign( {}, { kA, kB, kC } ); console.log(`new_obj =\n`, new_obj); // {kA: "111", kB: "222", kC: "333"}

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');