类型“ T”不能用于索引类型“类”

Question

我正在寻求为特定类制作Parameters的通用打字稿。

type Example<T> = Parameters<UserRepository[T]>[0]

但是我收到此错误：

  

类型'T'不能用于索引类型“ UserRepository”

也尝试过：

type Example<T extends string> = Parameters<UserRepository[T]>[0]

我目前有这个：

type Params = Parameters<UserRepository["members"]>[0] & 
              Parameters<UserRepository["contacts"]>[0] & 
              Parameters<UserRepository["user"]>[0]

我想创建两个泛型ClassMethod和ClassMethods。

type ClassMethod<C, M> = string
type Param = ClassMethod<UserRepository, 'members'>

或

type ClassMethod<T> = Parameters<UserRepository[T]>[0]
type Param = ClassMethod<'members'> // where the generic has `UserRepository `in it.

和

type ClassMethods<C, M> = string
type Params = ClassMethods<UserRepository, ['members', 'contacts', 'user']>

Answer 1

ClassMethod<'meow'>

第一个示例很简单，该类被嵌入d

type ClassMethod<T extends keyof UserRepository> = Parameters<UserRepository[T]>[0]

ClassMethod<UserRepository, 'meow'>

type ClassMethod<c extends any, T extends keyof c> = Parameters<c[T]>[0]

ClassMethods<UserRepository, 'meow' | 'woof'>

class UserRepository {
    meow (i: { catSound: string }) {

    }
    woof (i: { dogSound: string }) {

    }
}

// As usual, see https://stackoverflow.com/questions/50374908/transform-union-type-to-intersection-type
type UnionToIntersection<U> = 
  (U extends any ? (k: U)=>void : never) extends ((k: infer I)=>void) ? I : never

type ClassMethods<C, M extends keyof C> = UnionToIntersection<{
    [K in M]: C[K] extends ((...args: any[]) => any) ? Parameters<C[K]>[0] : never
}[M]>

type Parms = ClassMethods<UserRepository, 'meow' | 'woof'>

// Or if you really want an array...
type ClassMethods2<C, M extends Array<keyof C>> = UnionToIntersection<{
    [K in M[number]]: C[K] extends ((...args: any[]) => any) ? Parameters<C[K]>[0] : never
}[M[number]]>

type Parms2 = ClassMethods2<UserRepository, ['meow', 'woof']>

特别感谢typescript gitter上的Gerrit Birkeland。