多个订阅嵌套在一个订阅中

Question

我发现自己对尝试设置一个非常简单的 rxjs 订阅流程感到困惑。将多个不相关的订阅嵌套到另一个订阅中。

我在一个有角度的应用程序中，在进行其他订阅之前我需要填写一个主题。

这将是我想要实现的嵌套版本。

subject0.subscribe(a => {

    this.a = a;

    subject1.subscribe(x => {
        // Do some stuff that require this.a to exists
    });

    subject2.subscribe(y => {
        // Do some stuff that require this.a to exists
    });

});

我知道嵌套订阅不是一个好的做法，我尝试使用flatMaporconcatMap但并没有真正了解如何实现这一点。

Answer 1

将每个 Observable 的数据流分开总是一个好主意，这样您以后就可以轻松地将它们组合起来。

const first$ = this.http.get('one').pipe(
  shareReplay(1)
)

用于shareReplay使 Observable 变得热门，这样它就不会http.get('one')针对每个订阅进行调用。

const second$ = this.first$.pipe(
  flatMap(firstCallResult => this.http.post('second', firstCallResult))
);

const third$ = this.first$.pipe(
  flatMap(firstCallResult => this.http.post('third', firstCallResult))
);

之后，您可以订阅您需要的 Observables：

second$.subscribe(()=>{}) // in this case two requests will be sent - the first one (if there were no subscribes before) and the second one

third$.subscribe(() => {}) // only one request is sent - the first$ already has the response cached

如果您不想将first$的值存储在任何地方，只需将其转换为：

this.http.get('one').pipe(
  flatMap(firstCallResult => combineLatest([
    this.http.post('two', firstCallResult),
    this.http.post('three', firstCallResult)
  ])
).subscribe(([secondCallResult, thirdCallResult]) => {})

您也可以使用BehaviorSubject它来存储其中的值：

const behaviorSubject = new BehaviorSubject<string>(null); // using BehaviorSubject does not require you to subscribe to it (because it's a hot Observable)
const first$ = behaviorSubject.pipe(
  filter(Boolean), // to avoid emitting null at the beginning
  flatMap(subjectValue => this.http.get('one?' + subjectValue))
)

const second$ = first$.pipe(
  flatMap(firstRes => this.http.post('two', firstRes))
)

const third$ = first$.pipe(
  flatMap(()=>{...})
)

behaviorSubject.next('1') // second$ and third$ will emit new values
behaviorSubject.next('2') // second$ and third$ will emit the updated values again