讓潛在異步RxJS觀察到的肯定異步

Question

有潛在異步觀察到，即是這樣的：

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

potentiallyButNotNecessarilyAsyncObservable.subscribe(console.log); 

應該做異步觀察到。如果它已經是異步的，不要再拖延它，所以我不能做potentiallyButNotNecessarilyAsyncObservable.delay(0)。

這怎麼辦？

Answer 1

您可以通過使用調度程序來控制如何安排和執行工作。例如，您可以編寫.observeOn(Rx.Scheduler.async)，並且將使用setTimeout或setInterval而不是同步遞歸調用來安排工作。這是高級主題，如果你想更好地瞭解它我建議你讀this great documentation on Schedulers

這裏有兩個例子，第一個是在一個蜱執行，第二次在幾個：

const sync$ = Rx.Observable.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$.observeOn(Rx.Scheduler.async); 
 

 
// "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');

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