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_OPEN
、POPUP_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}
}
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/