如何避免具有逻辑和副作用的嵌套订阅

Question

大多数时候我可以避免嵌套订阅，但我不确定如何使用此代码来做到这一点：

const appleStuff = obtainAppleStuff();
// observeAll returns Observable<Apple[]>
appleService.observeAll().subscribe(apples => {
  let apple = apples.find(this.appleFilter);
  if(!apple){
    apple = appleService.create();
    apple.type = "Red";
    apple.size = 5;
    appleService.update(apple);
    if(apples.length !== 0){
      this.appleService.observeWormsOfApple(apples[0]).subscribe(worms => {
        appleService.linkWorms(worms, apple);
      });
    }
  }
  this.linkAppleStuff(appleStuff, apple);
});

理想情况下，我希望在 tap() 或单个非嵌套 subscribe() 中产生所有副作用，我该怎么做？

Answer 1

  const appleStuff = obtainAppleStuff();
    appleService
      .observeAll()
      .pipe(
        map(apples => {
          let apple = apples.find(this.appleFilter);
          if (!apple) {
            apple = appleService.create();
            apple.type = "Red";
            apple.size = 5;
          }
          return [apples, apple];
        }),
        tap(([, apple]) => {
          appleService.update(apple);
          this.linkAppleStuff(appleStuff, apple);
        }),
        filter(([apples]) => !!apples.length),
        switchMap(([apples, apple]) => this.appleService.observeWormsOfApple(apples[0]).pipe(withLatestFrom([apple])))
      )
      .subscribe(([worms, apple]) => {
        appleService.linkWorms(worms, apple);
      });