typescript - 如何使用重载定义函数以恰好具有 1 或 2 个参数,其类型取决于所使用的字符串文字

标签 typescript overloading typescript-generics

我想实现一个 emitEvent(event, extra?) 函数,该函数将受到已知字符串的字符串文字枚举的约束,例如 POPUP_OPENPOPUP_CLOSED 等。此函数接受第二个参数,它又是一个显式定义的字典形状,只能与特定事件键一起使用。

这是所有已知事件的字典:

interface Events {
    POPUP_OPEN: {name:'POPUP_OPEN',extra: {count:number}},
    POPUP_CLOSED: {name:'POPUP_CLOSED'},
    AD_BLOCKER_ON: {name:'AD_BLOCKER_ON', extra: {serviceName:string}},
    AD_BLOCKER_OFF: {name:'AD_BLOCKER_OFF'}
}

在要求的类型约束下使用:

// $ExpectType  {object: string; action: string; value: string;}
const t1 = emitEvent('POPUP_CLOSED')

// $ExpectError -> no extra argument allowed
const t11 = emitEvent('POPUP_CLOSED', {what:'bad'})

// $ExpectType  {object: string;action: string;foo: string; count: number;}  
const t2 = emitEvent('POPUP_OPEN',{count: 1231})

// $ExpectError -> extra argument is missing
const t22 = emitEvent('POPUP_OPEN')  

我的实现:

这个实现有一个大问题,对于字典值,它有extra定义,当它没有定义时,TS不会提示

// ✅ NO ERROR
const t2 = emitEvent('POPUP_OPEN',{count: 1231})
// ✅ Error
const t2 = emitEvent('POPUP_OPEN',{})
// 🚨NO ERROR -> THIS SHOULD ERROR !
const t2 = emitEvent('POPUP_OPEN')

实现:

type ParsedEvent<Extra = void> = Extra extends object ? BaseParsedEvents & Extra : BaseParsedEvents

type BaseParsedEvents = {
  object:string
  action:string
  value:string
}

type KnownEvents = keyof Events
type GetExtras<T> = T extends {name: infer N, extra: infer E} ? E : never;

function emitEvent<T extends KnownEvents>(event:T): ParsedEvent
function emitEvent<T extends KnownEvents, E extends GetExtras<Events[T]>>(event:T, extra: E): ParsedEvent<E>
function emitEvent<T extends string, E extends object>(event:T, extra?: E) {
  const parsedEmit = parse(event)
  return {...parsedEmit,...extra}
}


function parse(event:string): BaseParsedEvents {
    const [object,action,value] = event.split('_')  
    return {object,action,value}
}

总的来说,恐怕这是不可能实现的,但希望我是错的:)

最佳答案

您可以使用tuples in rest parameters而不是重载以使函数接受可变数量的参数。 GetExtras 将返回具有单个 E 元素的元组或空元组。然后我们可以传播 GetExtras 作为函数的传播参数:

interface Events {
    POPUP_OPEN: {name:'POPUP_OPEN',extra: {count:number}},
    POPUP_CLOSED: {name:'POPUP_CLOSED'},
    AD_BLOCKER_ON: {name:'AD_BLOCKER_ON', extra: {serviceName:string}},
    AD_BLOCKER_OFF: {name:'AD_BLOCKER_OFF'}
}


// $ExpectType  {object: string; action: string; value: string;}
const t1 = emitEvent('POPUP_CLOSED')

// ✅ NO ERROR
const t21 = emitEvent('POPUP_OPEN',{count: 1231})
// ✅ Error
const t213 = emitEvent('POPUP_OPEN',{})
// ✅  ERROR  as expected
const t223 = emitEvent('POPUP_OPEN')


type ParsedEvent<Extra extends [object] | [] = []> = Extra extends [infer E] ? BaseParsedEvents & E : BaseParsedEvents

type BaseParsedEvents = {
  object:string
  action:string
  value:string
}

type KnownEvents = keyof Events
type GetExtras<T> = T extends {name: infer N, extra: infer E} ? [E] : [];

function emitEvent<T extends KnownEvents, E extends GetExtras<Events[T]>>(event:T, ...extra: E): ParsedEvent<E>
function emitEvent<T extends string, E extends object>(event:T, extra?: E) {
  const parsedEmit = parse(event)
  return {...parsedEmit,...extra}
}


function parse(event:string): BaseParsedEvents {
    const [object,action,value] = event.split('_')  
    return {object,action,value}
}

play

关于typescript - 如何使用重载定义函数以恰好具有 1 或 2 个参数,其类型取决于所使用的字符串文字,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/57761604/

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