如何定义具有重载的函数以恰好具有 1 或 2 个参数,这些参数的类型取决于使用的字符串文字
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【中文标题】如何定义具有重载的函数以恰好具有 1 或 2 个参数,这些参数的类型取决于使用的字符串文字【英文标题】:How to define function with overloads to have exactly 1 or 2 arguments which types depend on used string literal 【发布时间】:2020-01-05 18:23:08 【问题描述】:我想实现一个emitEvent(event, extra?)
函数,该函数将受已知字符串的字符串字面值枚举约束,例如POPUP_OPEN
、POPUP_CLOSED
等。此函数接受第二个参数,它又是一个明确定义的字典形状,它只能与特定的事件键一起使用。
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'
要求类型约束的用法:
// $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,对于定义了extra
的字典值,没有定义时TS不会抱怨
// ✅ 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')
实施:
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,我担心这是不可能实现的,但希望我错了:)
【问题讨论】:
【参考方案1】:您可以使用tuples in rest parameters 代替重载来让函数接受可变数量的参数。 GetExtras
将返回一个带有单个 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
【讨论】:
不错!也可以通过此实现删除重载,但是通过删除,我们将不得不手动从其余数组中仅选择第一项,这确实不是很好的解决方案...ts function emitEvent<T extends KnownEvents, E extends GetExtras<Events[T]>>(event:T, ...extra: E) const parsedEmit = parse(event) return ...parsedEmit,...extra[0]
【参考方案2】:
如何将您的第一个函数重载声明限制为不包含extra
属性的事件?
// define events, that do not have extra property
// "POPUP_CLOSED" | "AD_BLOCKER_OFF"
type KnownEventsWithoutExtra =
[K in keyof Events]: Events[K] extends name: string; extra: object
? never
: K
[keyof Events];
// single argument overload is restricted to above events
function emitEvent<T extends KnownEventsWithoutExtra>(event: T): ParsedEvent;
// will error now
const t4 = emitEvent("POPUP_OPEN");
Playground
【讨论】:
我最初试图通过映射类型来做到这一点,但无法做到。有时我只是让我猜想的事情过于复杂:D 这绝对有效!尽管重载实现更通用,所以我将其标记为解决方案。谢谢!以上是关于如何定义具有重载的函数以恰好具有 1 或 2 个参数,这些参数的类型取决于使用的字符串文字的主要内容,如果未能解决你的问题,请参考以下文章