Typescript 使用动态输出类型和泛型可重用接口提取对象属性

很难 Select 一个适当的总结，所以让我来解释一下我的问题.我有一堆类，每个类都有一个toJSON方法，可以将类转换为它的JSON表示.我想 for each 方法添加一个新的可选参数，它允许我只返回类属性的一个子集，而不是所有属性.该方法的返回类型应该根据开发人员 Select 的属性进行相应的更改.

我想使此类型安全，但我还没有达到完全满足我的需要的程度.以下是缺少的简短示例:

interface IUser {
  id: string;
  name: string;
  age: number;
}

const fullUser = user.toJson(); // full "IUser"
const truncatedUser = user.toJson({ attributes: ['name'] }); // Pick<IUser, 'name'>

fullUser.id // works
truncatedUser.id // should throw an error, but tsc compiles

完整的例子，我已经实现了到目前为止.

// Using the generic here, because I want to pass interface IUser, or IABC, or IDEF etc
interface IJSONExportOptions<Type> {
  /** A list of object properties that are included in the the resulting JSON.  */
  attributes?: (keyof Type)[];
}

interface IUser {
  id: string;
  name: string;
  age: number;
}

class User {
  private id: string;
  private name: string;
  private age: number;

  constructor() {
    this.id = '1';
    this.name = 'Jon Doe';
    this.age = 20;
  }

  toJSON(opts?: IJSONExportOptions<IUser>) {
    const fullObj: IUser = {
      id: this.id,
      name: this.name,
      age: this.age,
    };

    if (!opts?.attributes?.length) {
      return fullObj as Pick<IUser, keyof IUser>;
    }

    const truncatedUser: Partial<Pick<IUser, keyof IUser>> = {};
    for (const key of opts?.attributes) {
      const val = fullObj[key];
      truncatedUser[key] = val;
    }
    return truncatedUser as Pick<IUser, keyof IUser>;
  }
}

const user = new User()
const fullUser = user.toJSON();

console.log(fullUser.age);

const truncatedUser = user.toJSON({ attributes: ['name'] });

console.log(truncatedUser.age); // I want that tsc throws an error here, but also that my IDE does not provide autocomplete for "undefined" properties

如果这很重要，代码使用typescript 4.4.3(升级是一个选项).

如有任何帮助，我们不胜感激.