「重学TS 2.0 」TS 练习题第四十七题

[semlinker]

实现 RequireExactlyOne 工具类型，用于满足以下功能。即只能包含 age 或 gender 属性，不能同时包含这两个属性。具体的使用示例如下所示：

interface Person {
  name: string;
  age?: number;
  gender?: number;
}

// 只能包含Keys中唯一的一个Key
type RequireExactlyOne<T, Keys extends keyof T> = // 你的实现代码

const p1: RequireExactlyOne<Person, 'age' | 'gender'> = {
  name: "lolo",
  age: 7,
};

const p2: RequireExactlyOne<Person, 'age' | 'gender'> = {
  name: "lolo",
  gender: 1
};

// Error
const p3: RequireExactlyOne<Person, 'age' | 'gender'> = {
  name: "lolo",
  age: 7,
  gender: 1
};

请在下面评论你的答案

[zhaoxiongfei]

interface Person {
  name: string;
  age?: number;
  gender?: number;
  income?: number;
}

// 只能包含Keys中唯一的一个Key
type RequireExactlyOne<T, Keys extends keyof T, K extends keyof T = Keys> = Keys extends any
  ? Omit<T, K> & Required<Pick<T, Keys>> & Partial<Record<Exclude<K, Keys>, never>>
  : never;

type T1 = RequireExactlyOne<Person, "age" | "gender" | "income">;
const p1: T1 = {
  name: "lolo",
  age: 7,
};

const p2: T1 = {
  name: "lolo",
  gender: 1
};

const p3: T1 = {
  name: "lolo",
  income: 100
};

// Error
const p4: T1 = {
  name: "lolo",
  age: 7,
  gender: 1,
};

const p5: T1 = {
  name: "lolo",
  age: 7,
  income: 1,
};

const p6: T1 = {
  name: "lolo",
  gender: 7,
  income: 1,
};

const p7: T1 = {
  name: "lolo",
  age: 8,
  gender: 7,
  income: 1,
};

解题思路: 利用联合类型 extends 实现分布执行，之后重点是 如何让联合类型规则只有其中某一个生效，在每一个上设置哪些禁止的属性为可选 never

[pagnkelly]

interface Person1 {
  name: string;
  age?: number;
  gender?: number;
}
// 想要构建成这个样子才可以满足条件
type ttt = { name: string } & ({ age: number, gender?: never } | { age?: never, gender: number })

type RP<T, K> = {
  [P in keyof T]: P extends K ? T & Partial<Record<P, never>> : never
}[keyof T];

// 只能包含Keys中唯一的一个Key
type RequireExactlyOne<T, K extends keyof T> = Omit<T, K> & RP<Pick<T, K>, K>

const p1: RequireExactlyOne<Person1, 'age' | 'gender'> = {
  name: "lolo",
  age: 7,
};

const p2: RequireExactlyOne<Person1, 'age' | 'gender'> = {
  name: "lolo",
  gender: 1
};

// Error
const p3: RequireExactlyOne<Person1, 'age' | 'gender'> = {
  name: "lolo",
  age: 7,
  gender: 1
};

[zjxxxxxxxxx]

type RequireExactlyOne<T, K extends keyof T> = K extends any
  ? Omit<T, K> & Partial<Record<K, never>>
  : never;

[bill-lai]

type RequireExactlyOne<T, Keys extends keyof T, K extends keyof T = Keys> = 
  Omit<T, K> & Partial<Record<K, never>> | (
    Keys extends any 
      ? Omit<T, K> & Required<Pick<T, Keys>> & Partial<Record<Exclude<K, Keys>, never>>
      : never 
  )

[dolphin0618]

type RequireExactlyOne<T, K extends keyof T, W extends keyof T = K> = Omit<T, W> & (
  K extends any ? Required<Pick<T, K>> & Partial<Record<Exclude<W, K>, undefined>> : 1
)

[zhengyimeng]

// 只能包含Keys中唯一的一个Key
type RequireExactlyOne<T, Keys extends keyof T> = Keys extends any
  ? Omit<T, Keys> & { [K in Keys]: never }
  : never;

为啥这样不对呢？

[zhengyimeng]

// 只能包含Keys中唯一的一个Key
type RequireExactlyOne<T, Keys extends keyof T> = Keys extends any
  ? Omit<T, Keys> & { [K in Keys]?: never }
  : never;

抱歉少了一个 可选