Question

在从map返回数组并将其提供给filter链中的rxjs时，我迷上了类型。所以这是一个例子：

export interface SunInterface {
  brightness: number;
}

export interface MoonInterface {
  people: string[];
}

export class HelloComponent  {

  constructor() {
    const first_obs$: Observable<SunInterface> = of({ brightness: 100 });
    const second_obs$: Observable<MoonInterface> = Observable.create((observer) => {
      setTimeout(() => {
      observer.next({ people: ['Neil Armstrong', 'Buzz Aldrin', 'Alan Bean'] });
      }, 2000)
    });

    combineLatest(
      first_obs$,
      second_obs$
    )
    .pipe(
      map(([sun, moon]) => {
        const filtred = moon.people.filter(astronaut => astronaut.startsWith('N'));
        return [sun, { ...moon, ...{ people: filtred }}]; // should I add type for arguments? 
      }),
      filter(([sun, moon]) => {
        console.log('1111', sun);
        console.log('2222', moon);
        return !!moon.people.length; // here I get a lint error
      })
    )
    .subscribe(([sun, moon]) => {
    })
  }
}

我在代码Property 'people' does not exist on type SunInterface | { people: string[] }中留下注释的地方出现了一个皮棉错误。我猜我在类型方面做错了。有人可以帮我吗？

Answer 1

尝试在管道中定义type

.....filter(([sun, moon]: [SunInterface, MoonInterface]) => ..... )

Rxjs：卡在返回参数数组中

1 个答案: