RxJS operator waitUntil
a: 1---2-3-4--5---6
b: ------T---------
o: ------1234-5---6
Using RxJS, is there some operator that can accomplish the diagram above? I have stream A which is a random stream of events, given a stream B which has a single true event, can I have an output stream that doesn't emit anything until that true event, and then sends everything is had saved up until then and afterwards emits normally?
I thought maybe I could use buffer(), but it seems like there is no way to do a one time buffer like this with that operator.
rxjs
asked Jan 18 at 18:52
delashumdelashum
706
add a comment |
a: 1---2-3-4--5---6
b: ------T---------
o: ------1234-5---6
Using RxJS, is there some operator that can accomplish the diagram above? I have stream A which is a random stream of events, given a stream B which has a single true event, can I have an output stream that doesn't emit anything until that true event, and then sends everything is had saved up until then and afterwards emits normally?
I thought maybe I could use buffer(), but it seems like there is no way to do a one time buffer like this with that operator.
rxjs
asked Jan 18 at 18:52
delashumdelashum
706
You can use delay operator for that
– Oles Savluk
Jan 19 at 9:10
add a comment |
a: 1---2-3-4--5---6
b: ------T---------
o: ------1234-5---6
Using RxJS, is there some operator that can accomplish the diagram above? I have stream A which is a random stream of events, given a stream B which has a single true event, can I have an output stream that doesn't emit anything until that true event, and then sends everything is had saved up until then and afterwards emits normally?
I thought maybe I could use buffer(), but it seems like there is no way to do a one time buffer like this with that operator.
rxjs
asked Jan 18 at 18:52
delashumdelashum
706
a: 1---2-3-4--5---6
b: ------T---------
o: ------1234-5---6
Using RxJS, is there some operator that can accomplish the diagram above? I have stream A which is a random stream of events, given a stream B which has a single true event, can I have an output stream that doesn't emit anything until that true event, and then sends everything is had saved up until then and afterwards emits normally?
I thought maybe I could use buffer(), but it seems like there is no way to do a one time buffer like this with that operator.
rxjs
rxjs
asked Jan 18 at 18:52
delashumdelashum
706
asked Jan 18 at 18:52
delashumdelashum
706
asked Jan 18 at 18:52
delashumdelashum
706
asked Jan 18 at 18:52
delashumdelashum
706
asked Jan 18 at 18:52
delashumdelashum
706
706
You can use delay operator for that
– Oles Savluk
Jan 19 at 9:10
add a comment |
You can use delay operator for that
– Oles Savluk
Jan 19 at 9:10
You can use delay operator for that
– Oles Savluk
Jan 19 at 9:10
You can use delay operator for that
– Oles Savluk
Jan 19 at 9:10
add a comment |
2 Answers
2
active
oldest
votes
I think @ZahiC's solution is correct but personally I'd do it in a single chain using the multicast operator.
a$.pipe(
multicast(new Subject(), s => concat(
s.pipe(
buffer(b$),
take(1),
),
s
)),
)
multicast will basically spit the stream into two where concat will first subscribe to the first one that is buffered until b$ emits. Then it completes immediately because of take(1) and concat subscribe to the same steam again but this time unbuffered.
answered Jan 20 at 12:14
martinmartin
43k1187130
add a comment |
const { concat, interval, of, from } = rxjs;
const { share, delay, toArray, takeUntil, mergeMap } = rxjs.operators;
const waitUntil = signal$ => source$ => {
const sharedSource$ = source$.pipe(share());
return concat(
sharedSource$.pipe(
takeUntil(signal$),
toArray(),
mergeMap(from)
),
sharedSource$
);
}
const stopWaiting$ = of('signal').pipe(delay(2000));
const source$ = interval(500).pipe(
waitUntil(stopWaiting$)
).subscribe(console.log);<script src="https://cdnjs.cloudflare.com/ajax/libs/rxjs/6.3.3/rxjs.umd.js"></script>answered Jan 19 at 10:44
ZahiCZahiC
3,92611221
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
I think @ZahiC's solution is correct but personally I'd do it in a single chain using the multicast operator.
a$.pipe(
multicast(new Subject(), s => concat(
s.pipe(
buffer(b$),
take(1),
),
s
)),
)
multicast will basically spit the stream into two where concat will first subscribe to the first one that is buffered until b$ emits. Then it completes immediately because of take(1) and concat subscribe to the same steam again but this time unbuffered.
answered Jan 20 at 12:14
martinmartin
43k1187130
add a comment |
I think @ZahiC's solution is correct but personally I'd do it in a single chain using the multicast operator.
a$.pipe(
multicast(new Subject(), s => concat(
s.pipe(
buffer(b$),
take(1),
),
s
)),
)
multicast will basically spit the stream into two where concat will first subscribe to the first one that is buffered until b$ emits. Then it completes immediately because of take(1) and concat subscribe to the same steam again but this time unbuffered.
answered Jan 20 at 12:14
martinmartin
43k1187130
add a comment |
I think @ZahiC's solution is correct but personally I'd do it in a single chain using the multicast operator.
a$.pipe(
multicast(new Subject(), s => concat(
s.pipe(
buffer(b$),
take(1),
),
s
)),
)
multicast will basically spit the stream into two where concat will first subscribe to the first one that is buffered until b$ emits. Then it completes immediately because of take(1) and concat subscribe to the same steam again but this time unbuffered.
answered Jan 20 at 12:14
martinmartin
43k1187130
I think @ZahiC's solution is correct but personally I'd do it in a single chain using the multicast operator.
a$.pipe(
multicast(new Subject(), s => concat(
s.pipe(
buffer(b$),
take(1),
),
s
)),
)
multicast will basically spit the stream into two where concat will first subscribe to the first one that is buffered until b$ emits. Then it completes immediately because of take(1) and concat subscribe to the same steam again but this time unbuffered.
answered Jan 20 at 12:14
martinmartin
43k1187130
answered Jan 20 at 12:14
martinmartin
43k1187130
answered Jan 20 at 12:14
martinmartin
43k1187130
answered Jan 20 at 12:14
martinmartin
43k1187130
43k1187130
add a comment |
add a comment |
const { concat, interval, of, from } = rxjs;
const { share, delay, toArray, takeUntil, mergeMap } = rxjs.operators;
const waitUntil = signal$ => source$ => {
const sharedSource$ = source$.pipe(share());
return concat(
sharedSource$.pipe(
takeUntil(signal$),
toArray(),
mergeMap(from)
),
sharedSource$
);
}
const stopWaiting$ = of('signal').pipe(delay(2000));
const source$ = interval(500).pipe(
waitUntil(stopWaiting$)
).subscribe(console.log);<script src="https://cdnjs.cloudflare.com/ajax/libs/rxjs/6.3.3/rxjs.umd.js"></script>answered Jan 19 at 10:44
ZahiCZahiC
3,92611221
add a comment |
const { concat, interval, of, from } = rxjs;
const { share, delay, toArray, takeUntil, mergeMap } = rxjs.operators;
const waitUntil = signal$ => source$ => {
const sharedSource$ = source$.pipe(share());
return concat(
sharedSource$.pipe(
takeUntil(signal$),
toArray(),
mergeMap(from)
),
sharedSource$
);
}
const stopWaiting$ = of('signal').pipe(delay(2000));
const source$ = interval(500).pipe(
waitUntil(stopWaiting$)
).subscribe(console.log);<script src="https://cdnjs.cloudflare.com/ajax/libs/rxjs/6.3.3/rxjs.umd.js"></script>answered Jan 19 at 10:44
ZahiCZahiC
3,92611221
add a comment |
const { concat, interval, of, from } = rxjs;
const { share, delay, toArray, takeUntil, mergeMap } = rxjs.operators;
const waitUntil = signal$ => source$ => {
const sharedSource$ = source$.pipe(share());
return concat(
sharedSource$.pipe(
takeUntil(signal$),
toArray(),
mergeMap(from)
),
sharedSource$
);
}
const stopWaiting$ = of('signal').pipe(delay(2000));
const source$ = interval(500).pipe(
waitUntil(stopWaiting$)
).subscribe(console.log);<script src="https://cdnjs.cloudflare.com/ajax/libs/rxjs/6.3.3/rxjs.umd.js"></script>answered Jan 19 at 10:44
ZahiCZahiC
3,92611221
const { concat, interval, of, from } = rxjs;
const { share, delay, toArray, takeUntil, mergeMap } = rxjs.operators;
const waitUntil = signal$ => source$ => {
const sharedSource$ = source$.pipe(share());
return concat(
sharedSource$.pipe(
takeUntil(signal$),
toArray(),
mergeMap(from)
),
sharedSource$
);
}
const stopWaiting$ = of('signal').pipe(delay(2000));
const source$ = interval(500).pipe(
waitUntil(stopWaiting$)
).subscribe(console.log);<script src="https://cdnjs.cloudflare.com/ajax/libs/rxjs/6.3.3/rxjs.umd.js"></script>const { concat, interval, of, from } = rxjs;
const { share, delay, toArray, takeUntil, mergeMap } = rxjs.operators;
const waitUntil = signal$ => source$ => {
const sharedSource$ = source$.pipe(share());
return concat(
sharedSource$.pipe(
takeUntil(signal$),
toArray(),
mergeMap(from)
),
sharedSource$
);
}
const stopWaiting$ = of('signal').pipe(delay(2000));
const source$ = interval(500).pipe(
waitUntil(stopWaiting$)
).subscribe(console.log);<script src="https://cdnjs.cloudflare.com/ajax/libs/rxjs/6.3.3/rxjs.umd.js"></script>const { concat, interval, of, from } = rxjs;
const { share, delay, toArray, takeUntil, mergeMap } = rxjs.operators;
const waitUntil = signal$ => source$ => {
const sharedSource$ = source$.pipe(share());
return concat(
sharedSource$.pipe(
takeUntil(signal$),
toArray(),
mergeMap(from)
),
sharedSource$
);
}
const stopWaiting$ = of('signal').pipe(delay(2000));
const source$ = interval(500).pipe(
waitUntil(stopWaiting$)
).subscribe(console.log);<script src="https://cdnjs.cloudflare.com/ajax/libs/rxjs/6.3.3/rxjs.umd.js"></script>answered Jan 19 at 10:44
ZahiCZahiC
3,92611221
edited Jan 19 at 12:33
answered Jan 19 at 10:44
ZahiCZahiC
3,92611221
answered Jan 19 at 10:44
ZahiCZahiC
3,92611221
answered Jan 19 at 10:44
ZahiCZahiC
3,92611221
3,92611221
add a comment |
add a comment |
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You can use delay operator for that
– Oles Savluk
Jan 19 at 9:10