TypeScript:通用接口作为其他接口的联合 [英] TypeScript: generic interface as union of other interfaces

问题描述

我想创建一个通用接口，其属性代表来自其他接口的属性的联合.

I would like to create a generic interface with properties that represent a union of properties from other interfaces.

假设我有两个接口

interface A {
    something: string;
    somethingElse: number;
}

interface B {
    something: Array<string>;
}

我不想把接口C写成

interface C {
    something: string | Array<string>;
    somethingElse?: number;
}

因为这意味着每当我修改接口 A 或 B 时，我也需要手动修改接口 C.

because that would mean that whenever I modify either of the interfaces A or B, I would need to manually modify interface C as well.

根据我在 TypeScript 文档中看到的以及 Stack Overflow 上的答案，我应该声明一个新类型

From what I've seen in the TypeScript documentation as well as answers here on Stack Overflow, I should declare a new type

type unionOfKeys = keyof A | keyof B;

并实现通用接口形式

interface GenericInterface {
    <T>(arg: T): T;
}

我正在考虑

interface C {
    <T extends unionOfKeys>(arg: T): T extends unionOfKeys ? A[T] | B[T] : any
}

但由于许多属性与其类型不匹配而失败.

but that fails because of mismatch between a number of properties and their types.

我将不胜感激.谢谢.

推荐答案

我认为以下版本的 MergeUnion 可能表现得如您所愿:

I think the following version of MergeUnion<T> might behave how you want:

type MergeUnion<T> = (
  keyof T extends infer K ? [K] extends [keyof T] ? Pick<T, K> & {
    [P in Exclude<(T extends any ? keyof T : never), K>]?:
    T extends Partial<Record<P, infer V>> ? V : never
  } : never : never
) extends infer U ? { [K in keyof U]: U[K] } : never;

type C = MergeUnion<A | B>;
// type C = { 
//  something: string | string[]; 
//  somethingElse?: number | undefined; }
// }

这类似于另一个答案，因为它找到了 T 的所有组成部分的所有键的并集(称之为 UnionKeys，定义为 Textends any ? keyof T : never) 并返回一个包含所有这些的映射类型.不同的是，这里我们还找到了T的所有成分的所有key的交集(称之为IntersectKeys，定义为只是keyof T) 并将键 T 拆分为两组键.来自交集的一个存在于每个组成部分中，因此我们只需执行 Pick 即可获取公共属性.余数 Exclude> 在最终类型中将是可选的.

This is similar to the other answer in that it finds the union of all keys of all the constituents of T (call it UnionKeys, defined as T extends any ? keyof T : never) and returns a mapped type with all of them in it. The difference is that here we also find the intersection of all keys of all the constituents of T (call it IntersectKeys, defined as just keyof T) and split the keys T into two sets of keys. The one from the intersection are present in every constituent, so we can just do Pick<T, IntesectKeys> to get the common properties. The remainder, Exclude<UnionKeys, IntersectKeys> will be optional in the final type.

UPDATE 2019-08-23:下面提到的错误似乎从 TS3.5.1 开始修复

UPDATE 2019-08-23: the bug mentioned below seems to be fixed as of TS3.5.1

它很丑，如果我感觉好点我会清理它.问题是当出现在所有成分中的任何属性本身都是可选的时，仍然存在问题.TypeScript 中的错误(从 TS3.5 开始)在 {a?: 字符串} |{a?: number}，a 属性被视为必需 属性，如 {a: string |数量 |undefined}，而如果任何组成部分将其视为可选，则将其视为可选会更正确.该错误渗透到 MergeUnion:

It's pretty ugly, and I'd clean it up if I felt better about it. The problem is that there's still an issue when any of the properties appearing in all constituents are themselves optional. There's a bug in TypeScript (as of TS3.5) where in {a?: string} | {a?: number}, the a property is seen as a required property like {a: string | number | undefined}, whereas it would be more correct to be treated as optional if any of the constituents have it as optional. That bug bleeds through to MergeUnion:

type Oops = MergeUnion<{a?: string} | {a?: number}>
// type Oops =  { a: string | number | undefined; }

我没有更好的答案而不是更复杂的，所以我就到此为止.

I don't have a great answer there that isn't even more complicated, so I'll stop here.

也许这足以满足您的需求.或者@TitianCernicova-Dragomir 的回答可能足以满足您的需求.希望这些答案对您有所帮助；祝你好运！

Maybe this is sufficient for your needs. Or maybe @TitianCernicova-Dragomir's answer is sufficient for your needs. Hope these answers help you; good luck!

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