typescript - 如何声明在类型别名中声明的仅一个别名的数组

Question

如果我有一个像这样定义的 TypeScript 类型别名:

export type PlatformType = 'em' | 'ea' | 'hi';

我如何在接口(interface)中声明一个属性，以便该属性是每个 PlatformType 的一种类型的数组别名？例如，我想要这个:

export interface Constants {
  platformTypes: PlatformType[]; // how to declare a type for platformTypes?
}

所以当我有一个 Constants 的对象时类型，我需要这个:

export const constants: Constants = {
  platformTypes: ['em', 'ea', 'hi']
}

基本上，如果我有这个，我想得到一个错误:

export const constants: Constants = {
  platformTypes: ['em', 'ea'] // should error, missing 'hi'
}

或者

export const constants: Constants = {
  // this will error anyway because 'extra' doesn't belong to PlatformType alias
  platformTypes: ['em', 'ea', 'hi', 'extra']
}

或者

export const constants: Constants = {
  platformTypes: ['em', 'em', 'ea', 'hi'] // should error, duplicate 'em'
}

Answer 1

我认为您应该定义所有允许的元组的排列。
例如，如果您有三个元素的联合，您需要创建至少 6 个元组:

["em", "ea", "hi"] | ["em", "hi", "ea"] | ["ea", "em", "hi"] | ["ea", "hi", "em"] | ["hi", "em", "ea"] | ["hi", "ea", "em"]

这是因为联合是无序的数据结构。

export type PlatformType = 'em' | 'ea' | 'hi';


// credits goes to https://twitter.com/WrocTypeScript/status/1306296710407352321
type TupleUnion<U extends string, R extends any[] = []> = {
    [S in U]: Exclude<U, S> extends never ? [...R, S] : TupleUnion<Exclude<U, S>, [...R, S]>;
}[U];

export interface Constants {
  platformTypes: TupleUnion<PlatformType>; // how to declare a type for platformTypes?
}

export const constants: Constants = {
  platformTypes: ['em', 'ea', 'hi'] // ok
}

export const constants2: Constants = {
  platformTypes: ['em', 'ea'] // error
}

export const constants3: Constants = {
  platformTypes: ['em', 'ea', 'hi', 'extra'] // error
}

export const constants4: Constants = {
  platformTypes: ['em', 'em', 'ea', 'hi'] // error
}

Playground
请记住， typescript 中联合的顺序很棘手。
请查看 this question和 this问题
您可以在我的 blog 中找到更多示例