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
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];
其他回答
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');
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 };
使用对象解构和展开运算符的变体:
const old_obj = {
k1: `111`,
k2: `222`,
k3: `333`
};
// destructuring, with renaming. The variable 'rest' will hold those values not assigned to kA, kB, or kC.
const {
k1: kA,
k2: kB,
k3: kC,
...rest
} = old_obj;
// now create a new object, with the renamed properties kA, kB, kC;
// spread the remaining original properties in the 'rest' variable
const newObj = {kA, kB, kC, ...rest};
对于一个键,这可以很简单:
const { k1: kA, ...rest } = old_obj;
const new_obj = { kA, ...rest }
你也可能喜欢更“传统”的风格:
const { k1, ...rest } = old_obj
const new_obj = { kA: k1, ...rest}
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);
我想这么做
const originalObj = {
a: 1,
b: 2,
c: 3, // need replace this 'c' key into 'd'
};
const { c, ...rest } = originalObj;
const newObj = { ...rest, d: c };
console.log({ originalObj, newObj });