RxJS Observables：为什么回调会触发两次？

Question

考虑这段代码，来自 Angular 6 个组件：

class AppComponent  {
  subject = new Subject<any>();
  one = this.subject.pipe(share(), tap(e => log('one emits')));
  two = this.one.pipe(tap(e => log('two emits')));
  three = this.one.pipe(delay(500), tap(e => log('three emits')));

  ngOnInit() {
    this.two.subscribe(e => log('two received'));
    this.three.subscribe(e => log('three received'));    
    this.subject.next();
  }  
}

当 ngOnInit 执行时，这是记录的内容：

one emits

two emits

two received

one emits

three emits

three received

我不明白：为什么one会发出两次？管道中的 share 运算符不应该使 two 和 three 订阅相同的共享源吗？

Source on Stackblitz

Answer 1

share() 运算符在您使用它时进行多播。所以当你在 tap 之前使用它时 tap 仍然有两个观察者。

所以只需在 tap 之后使用 share，它就会保持对其父级的订阅。

one = this.subject.pipe(tap(e => console.log('one emits')), share());

RxJS Observables：为什么回调会触发两次？

RxJS Observables: why does callback fire twice?

rxjs

angular

rxjs6