从类型联合定义打字稿函数签名[重复]答案

【问题标题】：Define typescript function signature from union of types [duplicate]从类型联合定义打字稿函数签名[重复]
【发布时间】：2021-01-07 08:45:14
【问题描述】：

我想从类型的联合中定义一个函数的类型：

type MyEvent =
| { type: 'hello', payload: {} }
| { type: 'start', payload: { date: Date } }


type On<Event> = Event extends { type: infer EventType, payload: infer EventPayload }
  ? (type: EventType, callback: (payload: EventPayload) => void) => void
  : never

const on: On<MyEvent>
on('hello', (payload) => void)

但编译器将第一个参数的类型定义为“从不”

On 返回一个看起来不错的类型。我也尝试了返回的原始类型，结果相同：

type RawOn =
| ((type: "hello", callback: (payload: {}) => void) => void)
| ((type: "start", callback: (payload: { date: Date }) => void) => void)

const raw_on: RawOn
raw_on('hello', (payload) => void)

如何定义“函数签名的联合类型”？

Playground

【问题讨论】：

标签： typescript generics types typescript-typings conditional-types

【解决方案1】：

感谢@cherryblossom 的解决方案：

type UnionToIntersection<U> = (U extends any ? (k: U) => void : never) extends ((k: infer I) => void)
  ? I
  : never

// --------------------------------------------------
  
type MyEvent =
| { type: 'hello', payload: {} }
| { type: 'start', payload: { date: Date } }


type On<Event> = UnionToIntersection<
  Event extends { type: infer EventType, payload: infer EventPayload }
    ? (type: EventType, callback: (payload: EventPayload) => void) => void
    : never
>

const init_on = (): On<MyEvent> => {}
const on = init_on()

// OK
on('hello', (payload: {}) => {})
on('start', (payload: { date: Date }) => {})

// ERROR (that is OK)
on('hello', (payload: { date: Date }) => {})
on('bad-event', (payload: {}) => {})

Playground

【讨论】：