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
在您最喜欢的编辑器中尝试一下
const obj = {1: 'a', 2: 'b', 3: 'c'}
const OLD_KEY = 1
const NEW_KEY = 10
const { [OLD_KEY]: replaceByKey, ...rest } = obj
const new_obj = {
...rest,
[NEW_KEY]: replaceByKey
}
其他回答
如果有人需要重命名object的键:
const renameKeyObject = (obj, oldKey, newKey) => { 如果 (旧键 === 新键) 返回 volj; Object.keys(obj).forEach((key) => { if (key === oldKey) { obj[newKey] = obj[key]; 删除 obj[键]; } else if (obj[key] !== null &&; typeof obj[key] === “object”) { obj[key] = renameKeyObject(obj[key], oldKey, newKey); } }); 返回卷; };
重命名键,但避免改变原始对象参数
oldJson=[{firstName:'s1',lastName:'v1'},
{firstName:'s2',lastName:'v2'},
{firstName:'s3',lastName:'v3'}]
newJson = oldJson.map(rec => {
return {
'Last Name': rec.lastName,
'First Name': rec.firstName,
}
})
output: [{Last Name:"v1",First Name:"s1"},
{Last Name:"v2",First Name:"s2"},
{Last Name:"v3",First Name:"s3"}]
最好有一个新的数组
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}
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');
下面是创建具有重命名键的新对象的示例。
let x = { id: "checkout", name: "git checkout", description: "checkout repository" };
let renamed = Object.entries(x).reduce((u, [n, v]) => {
u[`__${n}`] = v;
return u;
}, {});