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
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,
} */
我会这样做:
function renameKeys(dict, keyMap) {
return _.reduce(dict, function(newDict, val, oldKey) {
var newKey = keyMap[oldKey] || oldKey
newDict[newKey] = val
return newDict
}, {})
}
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];
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。
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');