Question

以下使用switchMap ...

Rx.Observable.timer(0, 1000)

  // First switchMap
  .switchMap(i => {
    if (i % 2 === 0) {
      console.log(i, 'is even, continue');
      return Rx.Observable.of(i);
    }
    console.log(i, 'is odd, cancel');
    return Rx.Observable.empty();
  })

  // Second switchMap
  .switchMap(i => Rx.Observable.timer(0, 500).mapTo(i))

  .subscribe(i => console.log(i));

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...产生以下输出：

0 is even, continue
0
0
1 is odd, cancel
0 // <-------------------- why?
0 // <-------------------- why?
2 is even, continue
2
2
3 is odd, cancel
2 // <-------------------- why?
2 // <-------------------- why?
4 is even, continue
.
.
.

我希望得到以下结果：

0 is even, continue
0
0
1 is odd, cancel
2 is even, continue
2
2
3 is odd, cancel
4 is even, continue
.
.
.

我预计第一个switchMap会在每次返回时取消所有下游，我猜这是不正确的。

有没有办法达到预期的行为？

Answer 1

原因是返回一个空的observable不会导致它切换到新的observable，也就是第二个switchMap永远不会被调用。

非常hacky 解决方案是返回一个你以后可以忽略的魔法值

const CANCEL = {};
Rx.Observable.timer(0, 1000)

  // First switchMap
  .switchMap(i => {
    if (i % 2 === 0) {
      console.log(i, 'is even, continue');
      return Rx.Observable.of(i);
    }
    console.log(i, 'is odd, send CANCEL observable');
    return Rx.Observable.of(CANCEL);
  })


  // Second switchMap
  .switchMap(i => Rx.Observable.timer(0, 500).mapTo(i))

  // Filter out cancelled events
  .filter(i => i != CANCEL)
  
  .subscribe(i => console.log(i));

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如何取消RxJS中的switchmap链？

1 个答案: