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 ];
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
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 };
其他回答
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;
}
}
);
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"}
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,
} */
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”……
重命名对象键的另一种方法:
让我们考虑这个对象:
let obj = {"name": "John", "id": 1, "last_name": "Doe"}
让我们重命名name key为first_name:
let { name: first_name, ...rest } = obj;
obj = { first_name, ...rest }
现在对象是:
{"first_name": "John", "id": 1, "last_name": "Doe"}