Question

我正在尝试为节点编写Google Maps帮助程序模块的声明，但我遇到了图书馆期望的PromiseConstructorLike问题，并将其返回＆＃34; PromiseLike＆＃34 ;实例方法正确（根据https://googlemaps.github.io/google-maps-services-js/docs/module-@google_maps.html）：

Promise     function    <optional>  Promise constructor (optional).

所以我做了（剥离了有趣的部分）：

declare namespace GoogleMaps {
  export interface CreateClientOptions<T> {
    /** Promise constructor (optional). */
    Promise?: T; 
  }

  export interface GoogleMapsClient<T> {
    directions<U>(query, callback?: ResponseCallback<U>): RequestHandle<U, T>;
  }

  export interface Response<U extends any> {
      headers: any;
      json: U;
      status: number;
  }

  export interface RequestHandle<U, T extends PromiseLike<Response<U>>> {
      asPromise(): T;
      cancel(): void;
      finally(callback: ResponseCallback<U>): void;
  }

  export type ResponseCallback<U> = (err: Error, result: Response<U>) => void; 
  export function createClient<T extends PromiseConstructorLike>(options: CreateClientOptions<T>): GoogleMapsClient<T>;
}

declare module '@google/maps' {
  export = GoogleMaps
}

当然，如果我使用createClient中的Bluebird作为

，它就不起作用了

import * as bluebird from 'bluebird'
import { createClient } from '@google/maps'

createClient({ Promise: bluebird }).directions({}).asPromise()/** no "then" here, just the static methods from Bluebird, like Bluebird.all */

问题是：

无论如何，我可以提示asPromise方法从bluebird返回实例方法（然后，catch，finally，reduce，timeout等），而不必扩展RequestHandle接口手动吗

更多信息（lib.d.ts声明）：

PromiseConstructorLike是：

 declare type PromiseConstructorLike = new <T>(executor: (resolve: (value?: T | PromiseLike<T>) => void, reject: (reason?: any) => void) => void) => PromiseLike<T>;

PromiseLike是：

interface PromiseLike<T> {
    /**
     * Attaches callbacks for the resolution and/or rejection of the Promise.
     * @param onfulfilled The callback to execute when the Promise is resolved.
     * @param onrejected The callback to execute when the Promise is rejected.
     * @returns A Promise for the completion of which ever callback is executed.
     */
    then(
        onfulfilled?: ((value: T) => T | PromiseLike<T>) | undefined | null,
        onrejected?: ((reason: any) => T | PromiseLike<T>) | undefined | null): PromiseLike<T>;
}

Answer 1

您的声明包含编译错误，该错误源于混淆Promise 实例类型和Promise 构造函数类型。 T中的类型参数GoogleMapsClient用于填充T中的RequestHandle，但在GoogleMapsClient中，这表示Promise构造函数类型，而在{ {1}}它代表RequestHandle实例类型。

您似乎打算根据Promise 实例类型Promise正确输入所有内容，其中PromiseLike<Response>是响应类型。但是，由于事先不知道U（即在致电U之前），因此遗憾的是不可能。

如果您想在GoogleMapsClient.directions之后致电then()，只需将asPromise()的返回类型更改为RequestHandle.asPromise，然后移除类型参数PromiseLike<Response>：

我个人还会在export interface RequestHandle { asPromise(): PromiseLike; cancel(): void; finally(callback: ResponseCallback): void; }和extends PromiseConstructorLike中添加约束T到参数CreateClientOptions，以便传递{{1}的类型安全性构造函数不仅仅依赖于GoogleMapsClient中指定的约束。

总而言之，声明现在如下所示：

Promise

通过这些声明，您的createClient示例有效，您可以在declare namespace GoogleMaps { export interface CreateClientOptions<T extends PromiseConstructorLike> { /** Promise constructor (optional). */ Promise?: T; } export interface GoogleMapsClient<T extends PromiseConstructorLike> { directions(query, callback?: ResponseCallback): RequestHandle; } export interface Response { headers: any; json: U; status: number; } export interface RequestHandle { asPromise(): PromiseLike<Response>; cancel(): void; finally(callback: ResponseCallback): void; } export type ResponseCallback = (err: Error, result: Response) => void; export function createClient<T extends PromiseConstructorLike>(options: CreateClientOptions<T>): GoogleMapsClient<T>; } declare module '@google/maps' { export = GoogleMaps }之后致电bluebird。

Answer 2

随着Typescript 2.8的发布，新的“推断”关键字使这成为可能！它可以推断（并传递）复杂的嵌套声明，解释器将尝试为您获取信息，从而提供非常好的强类型化体验。

所以，如果要获取构造函数的类型

    if (instance == null) {
    throw new IllegalStateException("this application does not 
    inherit GlobalApplication"); " +
    "}

    return instance;
    }

    Step4)
    g.setAdapter(new ImageAdapter(getGlobalApplicationContext()));

新的class MyPromise extends Promise<any> implements PromiseLike<any> { add(s: number) { s++ return this } dummy() { return this } } function typedFactory(u: U): InstanceType { return new u<void>(() => { }) as any // this isn't needed since we are just trying to show the functionality, // would be interfacing another library through types only, so that // the compiler doesn't b*tch about it } typedFactory(Promise).then(() => { }) typedFactory(MyPromise).add(1).dummy().then(() => {})实际上在InstanceType中可用，并且定义为：

lib.es5.d.ts

它显示了type InstanceType<T extends new (...args: any) => any> = T extends new (...args: any) => infer R ? R : any;关键字的真实功能，您可以在https://www.typescriptlang.org/play/中尝试

如何在接口声明中使用PromiseConstructorLike实例的原型？

2 个答案: