Question

可能存在异步的可观察性，例如：

const potentiallyButNotNecessarilyAsyncObservable = [
  Observable.of('sync'),
  Observable.of('async').delay(100)
][Math.round(Math.random())];

potentiallyButNotNecessarilyAsyncObservable.subscribe(console.log);

它应该是异步可观察的。如果它已经是异步的，那么我们不希望进一步延迟它，所以我不能potentiallyButNotNecessarilyAsyncObservable.delay(0)。

如何做到这一点？

Answer 1

您可以使用调度程序控制工作的安排和执行方式。例如，您可以编写.observeOn(Rx.Scheduler.async)并使用setTimeout或setInterval来安排工作，而不是同步递归调用。这是一个高级主题，如果您想更好地了解它，我建议您阅读this great documentation on Schedulers

以下是两个例子，第一个是在一个刻度中执行，第二个是在几个中执行：

const { of, asapScheduler } = rxjs; // = require("rxjs")
const { observeOn } = rxjs.operators; // = require("rxjs/operators")

const sync$ = of('sync');

// by default RxJS will pick a default scheduler by using the principle of least concurrency
// in this case it is a null or undefined Schedulr which means notifications are delivered synchronously and recursively
console.log('--------- default(null) Scheduler ---------------');
console.log('just before subscribe');
sync$.subscribe({
  next: x => console.log('got value ' + x),
  error: err => console.error('something wrong occurred: ' + err),
  complete: () => console.log('done'),
});
console.log('just after subscribe');


const async$ = sync$.pipe(
  observeOn(asapScheduler)
);

// "got value sync" is printed after "just after subscribe"
console.log('--------- Rx.Scheduler.async Scheduler ----------');
console.log('just before subscribe');
async$.subscribe({
  next: x => console.log('got value ' + x),
  error: err => console.error('something wrong occurred: ' + err),
  complete: () => console.log('done'),
});
console.log('just after subscribe');

<script src="https://unpkg.com/rxjs@6.2.2/bundles/rxjs.umd.min.js"></script>

使潜在的异步RxJS observable肯定是异步的

1 个答案: