问题
Is it possible to get all the "active" subscriptions without storing them manually?
I'd like to unsubscribe all of the "active" subscriptions and don't want to reference each of them in an array or a variable.
回答1:
I depends on whether you're using a Subject or an Observable but there's probably no way to do this "automatically".
Observables
I don't think you can have such thing as "subscribed Observable" because you either store an Observable or Subscription:
const source = Observable.of(...)
.map(...);
const subscription = source
.subscribe();
Here source represents an Observable and subscription represents a single subscription.
Note that you can have a Subscription instance that stores multiple other subscriptions:
const subscriptions = new Subscription();
const sub1 = Observable...subscribe();
const sub2 = Observable...subscribe();
const sub3 = Observable...subscribe();
subscriptions.add(sub1).add(sub2).add(sub3);
// Then unsubscribe all of them with a single
subscriptions.unsubscribe();
Subjects
If you're using Subjects they do have the unsubscribe method themselves, see https://github.com/ReactiveX/rxjs/blob/master/src/Subject.ts#L96.
However be aware that this makes the Subject "stopped", for more info see https://medium.com/@martin.sikora/rxjs-subjects-and-their-internal-state-7cfdee905156
回答2:
Yeap. Just call .observers property if you is using Subject object of the rxjs.
Hope this helps.
回答3:
One of the options is to use takeUntil in combination with Subject to stop all (Observable) subscriptions.
terminator$: Subject<boolean> = new Subject();
Observable.timer(1000)
.do(i => console.log(`timer 1: ${i}`))
.takeUntil(terminator$)
.subscribe();
Observable.timer(5000)
.do(i => console.log(`timer 2: ${i}`))
.takeUntil(terminator$)
.subscribe();
const stopSubscriptions = () => terminator$.next(true);
回答4:
I think the basic problem is that an Observable (with exception of Subject and derivatives) does not keep a reference to it's observers.
Without built-in references, you need to handle them externally in some form.
I think the best you could achieve is to create a reusable subscription 'override' to wrap the mechanism, although I doubt it's worth it.
import { Observable } from 'rxjs/Observable';
import { Subscription } from 'rxjs/Subscription';
const subscribeAndGuard = function(component, fnData, fnError = null, fnComplete = null) {
// Define the subscription
const sub: Subscription = this.subscribe(fnData, fnError, fnComplete);
// Wrap component's onDestroy
if (!component.ngOnDestroy) {
throw new Error('To use subscribeAndGuard, the component must implement ngOnDestroy');
}
const saved_OnDestroy = component.ngOnDestroy;
component.ngOnDestroy = () => {
console.log('subscribeAndGuard.onDestroy');
sub.unsubscribe();
// Note: need to put original back in place
// otherwise 'this' is undefined in component.ngOnDestroy
component.ngOnDestroy = saved_OnDestroy;
component.ngOnDestroy();
};
return sub;
};
// Create an Observable extension
Observable.prototype.subscribeAndGuard = subscribeAndGuard;
// Ref: https://www.typescriptlang.org/docs/handbook/declaration-merging.html
declare module 'rxjs/Observable' {
interface Observable<T> {
subscribeAndGuard: typeof subscribeAndGuard;
}
}
Ref this question Angular/RxJs When should I unsubscribe from Subscription
回答5:
You can just store one Subscription, and add to it.
private _subscriptions = new Subscription(); // yes you can do this!
Then add to it
_subscriptions.add(
this.layoutManager.width$.subscribe((w) => {
// any nested subscriptions can also use this mechanism
_subscriptions.add(...);
});
Then in ngOnDestroy() you just call _subscriptions.unsubscribe();.
I've been trying to make a 'clever' way to do this, but to be frank I've overengineered it quite a bit (including logging) and this is the simplest way I've come across.
回答6:
With this decorator you shouldn't put the unsubscribe(), this is done automatically.
For use this you only need declare onDestroy, you don't need to put anything in this.
It is very easy to use, I encourage you to use it.
Npm installation
Official documentation
来源:https://stackoverflow.com/questions/47040664/get-all-current-active-subscriptions