I want to reemit the last value of my observable at a fix interval, to I tried
obs.pipe(repeat({delay:1000})).subscribe(x => console.log('Emitted', x));
but it did not work. after looking into this, my observable is in fact a BehaviorSubject.
So my Question is Why does the 1st emits every second
of('Observable').pipe(repeat({ delay: 1000 })).subscribe(x => console.log(x));
but not the this?
var bs = new BehaviorSubject('BehaviorSubject');
bs.pipe(repeat({ delay: 1000 })).subscribe(x => console.log(x));
How to do it with my BehaviorSubject?
Edit
And I would also like to reset my timer when the subject emits a new value.
the solution I found is
var bs = new BehaviorSubject('BehaviorSubject');
bs.pipe(switchMap(x => timer(0,1000).pipe(map => () => x)).subscribe(x => console.log(x));
but it feels ugly.
You can derive an observable from your BehaviorSubject that switchMaps to a timer that emits the received value. Whenever the subject emits, the timer is reset and will emit the latest value:
const bs = new BehaviorSubject('initial value');
const repeated = bs.pipe(
switchMap(val => timer(0, 1000).pipe(
map(() => val)
))
);
Here's a StackBlitz demo.
So my Question is Why does the 1st emits every second, but not the this?
The reason your example code using of as the source works and not the code using the BehaviorSubject can be found in the documentation of the repeat operator:
Returns an Observable that will resubscribe to the source stream when the source stream completes.
The observable created using of completes after it emits the provided value, so it will resubscribe. Since the BehaviorSubject was not completed, it will not resubscribe.
I have the following
const timer1 = interval(1000).pipe(take(10));
const timer2 = interval(2000).pipe(take(6));
const merged = merge(timer1, timer2);
merged.subscribe(x => console.log(x));
Now this will only provide the value of the latest emitting observable.
I would like that, when one emit, I get the new value, and the latest from the other one.
I tried to use zip but it will not use the latest. Check the screenshot below for the missing value I wish to have. How can I achieve this ?
Join, would work, but join do not emit if the same observable emitted twice before the other one emitted a value.
You are looking for combineLatest.
combineLatest will emit whenever either of the source observables emit. Note: it will not emit for the first time until each observable has emitted at least once.
So with your example, it could look like this (StackBlitz):
const timer1 = interval(1000).pipe(take(10));
const timer2 = interval(2000).pipe(take(6));
const merged = combineLatest([timer1, timer2]);
merged.subscribe(([one, two]) => console.log(`${one}-${two}`));
The code
const a = new Rx.Subject().do(x => console.log('a'))
const b = a.mapTo(0)
const c = a.mapTo(1)
const d = Rx.Observable.merge(b, c)
d.subscribe(x => console.log('d'))
a.next(3)
And the output
a
d
a
d
Why does a got printed twice? Isn't Rx.Subject a hot observable?
The Subject itself is hot/shared.
However: Any(most!) operators that you append will create a new stream, with the previous stream(in this case the Subject) as source - the new stream, however, is (for most operators) not hot and will only be made hot by deriving a hot stream through appending a hot operator (like share or publish ect...)
So when you share your do, everything should work as expected.
const a = new Rx.Subject().do(x => console.log('a')).share();
const b = a.mapTo(0);
const c = a.mapTo(1);
const d = Rx.Observable.merge(b, c)
d.subscribe(x => console.log('d'));
a.next(3);
<script src="https://unpkg.com/rxjs/bundles/Rx.min.js"></script>
You need to understand cold/hot observable and subject.
A cold Observable is an Observable that re-executes its subscribe handler every time it's subscribed to:
const cold = new Observable(function subscribe(observer) {
console.log('subscribed');
observer.next(Math.random());
observer.complete();
});
// > subscribed
// sub 1: 0.1231231231231
cold.subscribe((num) => console.log('sub 1:', num));
// > subscribed
// sub 2: 0.09805969045
cold.subscribe((num) => console.log('sub 2:', num));
A hot Observable is a source Observable (cold or otherwise) that has a Subject between the source and subscribers. When a hot Observable is subscribed to, the subscription is internally routed to the inner Subject transparently, and the Subject is subscribed to the source Observable. This ensures the source Observable only has one subscriber (the Subject), and the Subject shares the source's value with many Subscribers:
const cold = new Observable(function subscribe(observer) {
console.log('subscribed');
observer.next(Math.random());
observer.complete();
});
const hot = cold.publish();
hot.subscribe((num) => console.log('sub 1:', num));
hot.subscribe((num) => console.log('sub 2:', num));
hot.connect(); // <-- this subscribes the inner Subject to the cold source
// > subscribed
// > sub 1: 0.249848935489
// > sub 2: 0.249848935489
You can make an Observable hot via multicast, which takes a function that returns a Subject to use when it's connected. There are also variants of multicast for convenience (such as publish) that create specific types of Subjects. publish() is a convenience method for multicast(() => new Subject())
In addition to connect(), which subscribes the inner Subject to the source and returns the underlying Subscription, you can call refCount(), which returns an Observable. When the Observable returned by refCount() is subscribed to once, it will automatically call connect() internally, and subsequent subscriptions won't reconnect. When all subscribers unsubscribe, refCount will automatically unsubscribe the inner Subject from the source. share() is a convenience method for source.publish().refCount().
So, it will work,
const a = new Rx.Subject().do(x => console.log('a')).share();
const b = a.mapTo(0);
const c = a.mapTo(1);
const d = Rx.Observable.merge(b, c)
d.subscribe(x => console.log('d'));
a.next(3);
I'm making use of the withLatestFrom operator in RxJS in the normal way:
var combined = source1.withLatestFrom(source2, source3);
...to actively collect the most recent emission from source2 and source3 and to emit all three value only when source1 emits.
But I cannot guarantee that source2 or source3 will have produced values before source1 produces a value. Instead I need to wait until all three sources produce at least one value each before letting withLatestFrom do its thing.
The contract needs to be: if source1 emits then combined will always eventually emit when the other sources finally produce. If source1 emits multiple times while waiting for the other sources we can use the latest value and discard the previous values. Edit: as a marble diagram:
--1------------2---- (source)
----a-----b--------- (other1)
------x-----y------- (other2)
------1ax------2by--
--1------------2---- (source)
------a---b--------- (other1)
--x---------y------- (other2)
------1ax------2by--
------1--------2---- (source)
----a-----b--------- (other1)
--x---------y------- (other2)
------1ax------2by--
I can make a custom operator for this, but I want to make sure I'm not missing an obvious way to do this using the vanilla operators. It feels almost like I want combineLatest for the initial emit and then to switch to withLatestFrom from then on but I haven't been able to figure out how to do that.
Edit: Full code example from final solution:
var Dispatcher = new Rx.Subject();
var source1 = Dispatcher.filter(x => x === 'foo');
var source2 = Dispatcher.filter(x => x === 'bar');
var source3 = Dispatcher.filter(x => x === 'baz');
var combined = source1.publish(function(s1) {
return source2.publish(function(s2) {
return source3.publish(function(s3) {
var cL = s1.combineLatest(s2, s3).take(1).do(() => console.log('cL'));
var wLF = s1.skip(1).withLatestFrom(s2, s3).do(() => console.log('wLF'));
return Rx.Observable.merge(cL, wLF);
});
});
});
var sub1 = combined.subscribe(x => console.log('x', x));
// These can arrive in any order
// and we can get multiple values from any one.
Dispatcher.onNext('foo');
Dispatcher.onNext('bar');
Dispatcher.onNext('foo');
Dispatcher.onNext('baz');
// combineLatest triggers once we have all values.
// cL
// x ["foo", "bar", "baz"]
// withLatestFrom takes over from there.
Dispatcher.onNext('foo');
Dispatcher.onNext('bar');
Dispatcher.onNext('foo');
// wLF
// x ["foo", "bar", "baz"]
// wLF
// x ["foo", "bar", "baz"]
I think the answer is more or less as you described, let the first value be a combineLatest, then switch to withLatestFrom. My JS is hazy, but I think it would look something like this:
var selector = function(x,y,z) {};
var combined = Rx.Observable.concat(
source1.combineLatest(source2, source3, selector).take(1),
source1.withLatestFrom(source2, source3, selector)
);
You should probably use publish to avoid multiple subscriptions, so that would look like this:
var combined = source1.publish(function(s1)
{
return source2.publish(function(s2)
{
return source3.publish(function(s3)
{
return Rx.Observable.concat(
s1.combineLatest(s2, s3, selector).take(1),
s1.withLatestFrom(s2, s3, selector)
);
});
});
});
or using arrow functions...
var combined = source1.publish(s1 => source2.publish(s2 => source3.publish(s3 =>
Rx.Observable.concat(
s1.combineLatest(s2, s3, selector).take(1),
s1.withLatestFrom(s2, s3, selector)
)
)));
EDIT:
I see the problem with concat, the withLatestFrom isn't getting the values. I think the following would work:
var combined = source1.publish(s1 => source2.publish(s2 => source3.publish(s3 =>
Rx.Observable.merge(
s1.combineLatest(s2, s3, selector).take(1),
s1.skip(1).withLatestFrom(s2, s3, selector)
)
)));
...so take one value using combineLatest, then get the rest using withLatestFrom.
I wasn't quite satisfied with the accepted answer, so I ended up finding another solution. Many ways to skin a cat!
My use-case involves just two streams - a "requests" stream and a "tokens" stream. I want requests to fire as soon as they are received, using the whatever the latest token is. If there is no token yet, then it should wait until the first token appears, and then fire off all the pending requests.
I wasn't quite satisfied with the accepted answer, so I ended up finding another solution. Essentially I split the request stream into two parts - before and after first token arrives. I buffer the first part, and then re-release everything in one go once I know that the token stream is non-empty.
const first = token$.first()
Rx.Observable.merge(
request$.buffer(first).mergeAll(),
request$.skipUntil(first)
)
.withLatestFrom(token$)
See it live here: https://rxviz.com/v/VOK2GEoX
For RxJs 7:
const first = token$.first()
merge(
request$.pipe(
buffer(first),
mergeAll()
),
request$.pipe(
skipUntil(first)
)
).pipe(
withLatestFrom(token$)
)
I had similar requirements but for just 2 observables.
I ended up using switchMap+first:
observable1
.switchMap(() => observable2.first(), (a, b) => [a, b])
.subscribe(([a, b]) => {...}));
So it:
waits until both observables emit some value
pulls the value from second observable only if the first one has changed (unlike combineLatest)
doesn't hang subscribed on second observable (because of .first())
In my case, second observable is a ReplaySubject. I'm not sure if it will work with other observable types.
I think that:
flatMap would probably work too
it might be possible to extend this approach to handle more than 2 observables
I was surprised that withLatestFrom will not wait on second observable.
In my mind, the most elegant way to achieve the different behavior of an existing RxJS operator is to wrap it into a custom operator. So that from the outside it looks just like any regular operator and doesn't require you to restructure your code each time you need this behavior.
Here is how you can create your own operator which behaves just like withLatestFrom, except that at the very beginning it will emit as soon as the first value of the target observable is emitted (unlike standard withLatestFrom, which will ignore the first emission of the source if the target hasn't yet emitted once). Let's call it delayedWithLatestFrom.
Note that it's written in TypeScript, but you can easily transform it to plain JS. Also, it's a simple version that supports only one target observable and no selector function - you can extend it as needed from here.
export function delayedWithLatestFrom<T, N>(
target$: Observable<N>
): OperatorFunction<T, [T, N]> {
// special value to avoid accidental match with values that could originate from target$
const uniqueSymbol = Symbol('withLatestFromIgnore');
return pipe(
// emit as soon target observable emits the first value
combineLatestWith<T, [N]>(target$.pipe(first())),
// skip the first emission because it's handled above, and then continue like a normal `withLatestFrom` operator
withLatestFrom(target$.pipe(skip(1), startWith(uniqueSymbol))),
map(([[rest, combineLatestValue], withLatestValue]) => {
// take combineLatestValue for the first time, and then always take withLatestValue
const appendedValue =
withLatestValue === uniqueSymbol ? combineLatestValue : withLatestValue;
return [rest, appendedValue];
})
);
}
// SAMPLE USAGE
source$.pipe(
delayedWithLatestFrom(target$)
).subscribe(console.log);
So if you compare it with the original marble diagram for withLatestFrom, it will differ only in one fact: while withLatestFrom ignores the first emissions and produces b1 as the first value, the delayedWithlatestFrom operator will emit one more value a1 at the beginning, as soon as the second observable emits 1.
a) Standard withLatestFrom:
b) Custom delayedWithLatestFrom:
Use combineLatest and filter to remove tuples before first full set is found then set a variable to stop filtering. The variable can be within the scope of a wrapping defer to do things properly (support resubscription). Here it is in java (but the same operators exist in RxJs):
Observable.defer(
boolean emittedOne = false;
return Observable.combineLatest(s1, s2, s3, selector)
.filter(x -> {
if (emittedOne)
return true;
else {
if (hasAll(x)) {
emittedOne = true;
return true;
} else
return false;
}
});
)
I wanted a version where tokens are fetched regularly - and where I want to retry the main data post on (network) failure. I found shareReplay to be the key. The first mergeWith creates a "muted" stream, which causes the first token to be fetched immediately, not when the first action arrives. In the unlikely event that the first token will still not be available in time, the logic also has a startWith with an invalid value. This causes the retry logic to pause and try again. (Some/map is just a Maybe-monad):
Some(fetchToken$.pipe(shareReplay({refCount: false, bufferSize: 1})))
.map(fetchToken$ =>
actions$.pipe(
// This line is just for starting the loadToken loop immediately, not waiting until first write arrives.
mergeWith(fetchToken$.pipe(map(() => true), catchError(() => of(false)), tap(x => loggers.info(`New token received, success: ${x}`)), mergeMap(() => of()))),
concatMap(action =>
of(action).pipe(
withLatestFrom(fetchToken$.pipe(startWith(""))),
mergeMap(([x, token]) => (!token ? throwError(() => "Token not ready") : of([x, token] as const))),
mergeMap(([{sessionId, visitId, events, eventIds}, token]) => writer(sessionId, visitId, events, token).pipe(map(() => <ISessionEventIdPair>{sessionId, eventIds}))),
retryWhen(errors =>
errors.pipe(
tap(err => loggers.warn(`Error writing data to WG; ${err?.message || err}`)),
mergeMap((_error: any, attemptIdx) => (attemptIdx >= retryPolicy.retryCount ? throwError(() => Error("It's enough now, already")) : of(attemptIdx))), // error?.response?.status (int, response code) error.code === "ENOTFOUND" / isAxiosError: true / response === undefined
delayWhen(attempt => timer(attempt < 2 ? retryPolicy.shortRetry : retryPolicy.longRetry, scheduler))
)
)
)
),
)
)
Thanks to everyone on this question-page for good inputs.
Based on the answer from #cjol
Here's a RxJs 7 implementation of a waitFor operator that will buffer the source stream until all input observables have emitted values, then emit all buffered events on the source stream. Any subsequent events on the source stream are emitted immediately.
// Copied from the definition of withLatestFrom() operator.
export function waitFor<T, O extends unknown[]>(
inputs: [...ObservableInputTuple<O>]
): OperatorFunction<T, [T, ...O]>;
/**
* Buffers the source until every observable in "from" have emitted a value. Then
* emit all buffered source values with the latest values of the "from" array.
* Any source events are emitted immediately after that.
* #param from Array of observables to wait for.
* #returns Observable that emits an array that concatenates the source and the observables to wait.
*/
export function waitFor(
from: Observable<unknown>[]
): (source$: Observable<unknown>) => Observable<unknown> {
const combined$ = combineLatest(from);
// This served as a conditional that switched on and off the streams that
// wait for the the other observables, or emits the source right away because
// the other observables have emitted.
const firstCombined$ = combined$.pipe(first());
return function (source$: Observable<unknown>): Observable<unknown> {
return merge(
// This stream will buffer the source until the other observables have all emitted.
source$.pipe(
takeUntil(firstCombined$), // without this it continues to buffer new values forever
buffer(firstCombined$),
mergeAll()
),
// This stream emits the source straight away and will take over when the other
// observables have emitted.
source$.pipe(skipUntil(firstCombined$))
).pipe(
withLatestFrom(combined$),
// Flatten it to behave like withLatestFrom() operator.
map(([source, combined]) => [source, ...combined])
);
};
}
All of the above solutions are not really on the point, therefore I made my own. Hope it helps someone out.
import {
combineLatest,
take,
map,
ObservableInputTuple,
OperatorFunction,
pipe,
switchMap
} from 'rxjs';
/**
* ### Description
* Works similar to {#link withLatestFrom} with the main difference that it awaits the observables.
* When all observables can emit at least one value, then takes the latest state of all observables and proceeds execution of the pipe.
* Will execute this pipe only once and will only retrigger pipe execution if source observable emits a new value.
*
* ### Example
* ```ts
* import { BehaviorSubject } from 'rxjs';
* import { awaitLatestFrom } from './await-latest-from.ts';
*
* const myNumber$ = new BehaviorSubject<number>(1);
* const myString$ = new BehaviorSubject<string>("Some text.");
* const myBoolean$ = new BehaviorSubject<boolean>(true);
*
* myNumber$.pipe(
* awaitLatestFrom([myString$, myBoolean$])
* ).subscribe(([myNumber, myString, myBoolean]) => {});
* ```
* ### Additional
* #param observables - the observables of which the latest value will be taken when all of them have a value.
* #returns a tuple which contains the source value as well as the values of the observables which are passed as input.
*/
export function awaitLatestFrom<T, O extends unknown[]>(
observables: [...ObservableInputTuple<O>]
): OperatorFunction<T, [T, ...O]> {
return pipe(
switchMap((sourceValue) =>
combineLatest(observables).pipe(
take(1),
map((values) => [sourceValue, ...values] as unknown as [T, ...O])
)
)
);
}
Actually withLatestFrom already
waits for every source
emits only when source1 emits
remembers only the last source1-message while the other sources are yet to start
// when source 1 emits the others have emitted already
var source1 = Rx.Observable.interval(500).take(7)
var source2 = Rx.Observable.interval(100, 300).take(10)
var source3 = Rx.Observable.interval(200).take(10)
var selector = (a,b,c) => [a,b,c]
source1
.withLatestFrom(source2, source3, selector)
.subscribe()
vs
// source1 emits first, withLatestFrom discards 1 value from source1
var source1 = Rx.Observable.interval(500).take(7)
var source2 = Rx.Observable.interval(1000, 300).take(10)
var source3 = Rx.Observable.interval(2000).take(10)
var selector = (a,b,c) => [a,b,c]
source1
.withLatestFrom(source2, source3, selector)
.subscribe()