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「重学TS 2.0 」TS 练习题第四十七题
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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
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
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
};
type RequireExactlyOne<T, K extends keyof T> = K extends any
? Omit<T, K> & Partial<Record<K, never>>
: never;
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
)
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
)
// 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?
// 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