如何在TypeScript中根据对象传入的key返回这些key的推断？

Question

typescript，函数如何根据传入的key，返回对象对应的这些key？比如：

const obj = {    name: 'Marcelo',    age: 27,    role: ['admin']};type ObjType = {    [key in keyof typeof obj]: typeof obj[key]}const getAttr = <T extends ObjType, K extends keyof T, KS extends Array<K>>(...attribute: KS): T[KS] => {    const result = {};    for (const attr of attribute) {        result[attr] = obj[attr];    }    return result;};const attr = getAttr('age', 'name');

现在这个写法肯定不对，变量attr的推断是unknown，如何写函数getAttr的返回定义，才能让变量attr的推断为{age:number, name:string },就是函数传入了哪些key，就只返回这些key的推断

Answer 1

const obj = {    name: 'Marcelo',    age: 27,    role: ['admin']};type ObjType = typeof objtype ObjKeyType = keyof ObjTypeconst getAttr = <T extends ObjKeyType>(...attributes: T[]): { [K in T]: ObjType[K] } => {  const result = {} as { [K in T]: ObjType[K] };  for (const attr of attributes) {    result[attr] = obj[attr];  }  return result;};const attr = getAttr('age', 'name');

Answer 2

想到了，使用Pick就可以了，如下：

const obj = {    name: 'Marcelo',    age: 27,    role: ['admin']};type ObjType = {    [key in keyof typeof obj]: typeof obj[key]}const getAttr = <T extends ObjType, K extends keyof T>(...attribute: Array<K>): Pick<T, Array<K>[number]> => {    const result = {};    for (const attr of attribute) {        // eslint-disable-next-line @typescript-eslint/ban-ts-comment        // @ts-ignore        result[attr] = obj[attr];    }    return result as Pick<T, Array<K>[number]>;};const attr = getAttr('age', 'name');

有更好的答案，欢迎补充！