rxjs - 将 WithLatestFrom 的值传递给另一个运算符 - Rxjs、Ngrx 效果

Question

我已经获得了此 ngrx 效果，并且我想通过 withLatestFrom 使用我所在州的数据。

$loadProduct = createEffect(() =>
  this.actions$.pipe(
    ofType(fromActions.loadProduct),
    withLatestFrom(this.store.select(fromSelectors.debounceTime)),
    delay(debounceTime), // how can I get the debounceTime value up here?
    map(_ => fromActions.foo())
  )
)

如何在延迟范围内得到这个debounceTime值？

Answer 1

因为延迟只是delayWhen(() => timer(duration)); ，我认为你可以使用delayWhen来解决问题:

$loadProduct = createEffect(() =>
  this.actions$.pipe(
    ofType(fromActions.loadProduct),
    withLatestFrom(this.store.select(fromSelectors.debounceTime)),
    
    delayWhen(([, debounceTimeVal]) => timer(debounceTimeVal)),

    map(_ => fromActions.foo())
  )
)

作为旁注，delayWhen 只是 an interesting use case of mergeMap :不发出内部值，而是发出外部值。