从 ngrx/store 效果返回 forkJoin 的结果

Returning the result of a forkJoin from an ngrx/store effect

我正在尝试使用 ngrx 存储 return forkJoin 的结果,就像这个伪代码演示的那样:

@Effect()
someEffect$: Observable<Action> = this.action$.pipe(
    ofType<SomeActionType>(ActionTypes.SomeActionType),
    switchMap((action: any) => 
        this.http.get<any>('some/url').pipe(
            map(someResult => {
                // this implementation is unimportant, just the gist of the flow I'm after
                const potentialResults = oneOrMany(someResult);

                if( potentialResults.length === 1 )   {
                    return new SomeAction(potentialResults[0]);
                } else {
                    observables: Observable<any> = getObservables(someResult);

                    forkJoin(observables).subscribe((result) =>
                        // this is where I get stuck    
                        return new SomeAction(result);
                    )
                }
            }
        ))
)

如何从 forkJoin 的结果同步 return 一个动作?目前,我正在直接向 forkJoin 块内的商店发送一个动作,但这很臭,我想知道我如何 return forkJoin 中的这个动作] 块,使用另一个运算符,例如 map 或类似的东西。有任何想法吗?

您不能 return 来自 map() 回调的 Observable。您需要使用 switchMap()(或另一个 xxxMap())来做到这一点。您也无法订阅 forkJoin 可观察对象。相反,您必须 map():

someEffect$: Observable<Action> = this.action$.pipe(
    ofType<SomeActionType>(ActionTypes.SomeActionType),
    switchMap(() => this.http.get<any>('some/url'),
    switchMap(someResult => {
        const potentialResults = oneOrMany(someResult);
        if (potentialResults.length === 1)   {
            return of(new SomeAction(potentialResults[0]));
        } else {
            const observables: Array<Observable<any>> = getObservables(someResult);
            return forkJoin(observables).map(result => new SomeAction(result))
        }
    })
)

您可以创建一个主题并return它:

someEffect$ = createEffect(() => this.actions$.pipe(
ofType(SomeType.Save),
mergeMap(action => {
  const obsList: Observable<any>[] = createList();

  const sub: Subject<Action> = new Subject<Action>();
  forkJoin(obsList).subscribe(() => {
    sub.next(new SomeAction());
    sub.complete();
  });
  return sub;
})));