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

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,
} */

其他回答

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'}

在寻找了很多答案后,这是我最好的解决方案:

const renameKey = (oldKey, newKey) => {
  _.reduce(obj, (newObj, value, key) => {
    newObj[oldKey === key ? newKey : key] = value
    return newObj
  }, {})
}

很明显,它没有替换原来的键,而是构造了一个新对象。 问题中的方法有效,但会改变对象的顺序,因为它将新的键-值添加到最后一个对象上。

您可以将工作包装在一个函数中,并将其分配给Object原型。也许可以使用流畅的界面样式使多个重命名流动。

Object.prototype.renameProperty = function (oldName, newName) {
     // Do nothing if the names are the same
     if (oldName === newName) {
         return this;
     }
    // Check for the old property name to avoid a ReferenceError in strict mode.
    if (this.hasOwnProperty(oldName)) {
        this[newName] = this[oldName];
        delete this[oldName];
    }
    return this;
};

ECMAScript 5 Specific

我希望语法不是这么复杂,但它肯定是很好的有更多的控制。

Object.defineProperty(
    Object.prototype, 
    'renameProperty',
    {
        writable : false, // Cannot alter this property
        enumerable : false, // Will not show up in a for-in loop.
        configurable : false, // Cannot be deleted via the delete operator
        value : function (oldName, newName) {
            // Do nothing if the names are the same
            if (oldName === newName) {
                return this;
            }
            // Check for the old property name to 
            // avoid a ReferenceError in strict mode.
            if (this.hasOwnProperty(oldName)) {
                this[newName] = this[oldName];
                delete this[oldName];
            }
            return this;
        }
    }
);

在我看来,你的方法是最优化的。但你最终会得到重新排序的密钥。新创建的密钥将附加在末尾。我知道你不应该依赖键的顺序,但如果你需要保存它,你将需要遍历所有键并一个接一个地构造新对象,在这个过程中替换有问题的键。

是这样的:

var new_o={};
for (var i in o)
{
   if (i==old_key) new_o[new_key]=o[old_key];
   else new_o[i]=o[i];
}
o=new_o;

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}