Question

我与Typescript的斗争是要获得作为父类的子类的类函数的返回类型。我所能实现的就是从父类获取返回类型。

我在最后添加了一段代码，并附有解释性评论。

export interface StepResult<T> {
  data: T;
}

export interface GetCompanyExecute {
  a: number;
}

class BaseStep {
  public async execute(): Promise<StepResult<any>> {
    return { data: {} }
  }
}

class GetCompany extends BaseStep {
  public async execute(): Promise<StepResult<GetCompanyExecute>> {
    return { data: { a: 1 } }
  }
}

// T from Promise<T>
type ThenArg<T> = T extends Promise<infer U> ? U :
  T extends (...args: any[]) => Promise<infer L> ? L :
  T

const getData = async <T extends BaseStep>(StepClass: typeof BaseStep): Promise<ThenArg<T['execute']>['data']> => {
  return 1 as any;
}

(async () => {
  // TODO: I need to get "GetCompanyExecute" here as returned type
  // but it returns any which is return type of BaseStep
  const a = await getData(GetCompany)

  // This is return type I want `await getData(GetCompany)` to be:
  type X = ThenArg<GetCompany['execute']>['data']  
})

Answer 1

const getData = async <T extends BaseStep>(StepClass: new () => T): Promise<ThenArg<T['execute']>['data']> => {
  return 1;
}

(async () => {
  const a = await getData(GetCompany) // typeof a === GetCompanyExecute
})

希望这可以为您提供帮助

在打字稿中获取子类的返回类型

1 个答案: