RxJs - forkJoin with empty array
I'm currently using forkJoin to wait for an array of Observable(s) to finish before pipe(ing) and tap(ping).
I noticed if the array is empty nothing is emitted and I cannot even tap. How do I solve this kind of problem? Should I just check if the array is empty?
myFirstFunction(...) {
const observables = ...
return forkJoin(observables)
}
mySecondFunction(...) {
return myFirstFunction().pipe(tap(() => ...))
}
rxjs
asked Nov 26 '18 at 9:09
LppEddLppEdd
9,09621647
add a comment |
I'm currently using forkJoin to wait for an array of Observable(s) to finish before pipe(ing) and tap(ping).
I noticed if the array is empty nothing is emitted and I cannot even tap. How do I solve this kind of problem? Should I just check if the array is empty?
myFirstFunction(...) {
const observables = ...
return forkJoin(observables)
}
mySecondFunction(...) {
return myFirstFunction().pipe(tap(() => ...))
}
rxjs
asked Nov 26 '18 at 9:09
LppEddLppEdd
9,09621647
add a comment |
I'm currently using forkJoin to wait for an array of Observable(s) to finish before pipe(ing) and tap(ping).
I noticed if the array is empty nothing is emitted and I cannot even tap. How do I solve this kind of problem? Should I just check if the array is empty?
myFirstFunction(...) {
const observables = ...
return forkJoin(observables)
}
mySecondFunction(...) {
return myFirstFunction().pipe(tap(() => ...))
}
rxjs
asked Nov 26 '18 at 9:09
LppEddLppEdd
9,09621647
I'm currently using forkJoin to wait for an array of Observable(s) to finish before pipe(ing) and tap(ping).
I noticed if the array is empty nothing is emitted and I cannot even tap. How do I solve this kind of problem? Should I just check if the array is empty?
myFirstFunction(...) {
const observables = ...
return forkJoin(observables)
}
mySecondFunction(...) {
return myFirstFunction().pipe(tap(() => ...))
}
rxjs
rxjs
asked Nov 26 '18 at 9:09
LppEddLppEdd
9,09621647
asked Nov 26 '18 at 9:09
LppEddLppEdd
9,09621647
asked Nov 26 '18 at 9:09
LppEddLppEdd
9,09621647
asked Nov 26 '18 at 9:09
LppEddLppEdd
9,09621647
asked Nov 26 '18 at 9:09
LppEddLppEdd
9,09621647
9,09621647
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
That's because forkJoin requires all source Observables to emit at least one item and when there are no source Observables there's nothing to emit. However, forkJoin will still send the complete notification so you can use for example defaultIfEmpty operator to make sure it always emits at least one next.
forkJoin(observables).pipe(
defaultIfEmpty(null),
).subscribe(...);
Demo: https://stackblitz.com/edit/rxjs-kkd1qa?file=index.ts
answered Nov 26 '18 at 9:19
martinmartin
46.5k1193137
Thanks Martin. Is the "will still send the complete notification" related only to theforkJoinoperator or is it more general?
– LppEdd
Nov 26 '18 at 9:24
That's related toforkJoin's internal functionality.
– martin
Nov 26 '18 at 9:24
I asked because I was looking at this page learnrxjs.io/operators/combination/forkjoin.html and found nothing mentioning this. Oh btw, had to usedefaultIfEmpty()
– LppEdd
Nov 26 '18 at 9:26
1
This is as edge case so I'm not surprised it's not mentioned there. I don't think this behavior is mentioned not even in the official documentation. You can see that it really works here stackblitz.com/edit/rxjs-kkd1qa?file=index.ts
– martin
Nov 26 '18 at 9:31
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
That's because forkJoin requires all source Observables to emit at least one item and when there are no source Observables there's nothing to emit. However, forkJoin will still send the complete notification so you can use for example defaultIfEmpty operator to make sure it always emits at least one next.
forkJoin(observables).pipe(
defaultIfEmpty(null),
).subscribe(...);
Demo: https://stackblitz.com/edit/rxjs-kkd1qa?file=index.ts
answered Nov 26 '18 at 9:19
martinmartin
46.5k1193137
Thanks Martin. Is the "will still send the complete notification" related only to theforkJoinoperator or is it more general?
– LppEdd
Nov 26 '18 at 9:24
That's related toforkJoin's internal functionality.
– martin
Nov 26 '18 at 9:24
I asked because I was looking at this page learnrxjs.io/operators/combination/forkjoin.html and found nothing mentioning this. Oh btw, had to usedefaultIfEmpty()
– LppEdd
Nov 26 '18 at 9:26
1
This is as edge case so I'm not surprised it's not mentioned there. I don't think this behavior is mentioned not even in the official documentation. You can see that it really works here stackblitz.com/edit/rxjs-kkd1qa?file=index.ts
– martin
Nov 26 '18 at 9:31
add a comment |
That's because forkJoin requires all source Observables to emit at least one item and when there are no source Observables there's nothing to emit. However, forkJoin will still send the complete notification so you can use for example defaultIfEmpty operator to make sure it always emits at least one next.
forkJoin(observables).pipe(
defaultIfEmpty(null),
).subscribe(...);
Demo: https://stackblitz.com/edit/rxjs-kkd1qa?file=index.ts
answered Nov 26 '18 at 9:19
martinmartin
46.5k1193137
Thanks Martin. Is the "will still send the complete notification" related only to theforkJoinoperator or is it more general?
– LppEdd
Nov 26 '18 at 9:24
That's related toforkJoin's internal functionality.
– martin
Nov 26 '18 at 9:24
I asked because I was looking at this page learnrxjs.io/operators/combination/forkjoin.html and found nothing mentioning this. Oh btw, had to usedefaultIfEmpty()
– LppEdd
Nov 26 '18 at 9:26
1
This is as edge case so I'm not surprised it's not mentioned there. I don't think this behavior is mentioned not even in the official documentation. You can see that it really works here stackblitz.com/edit/rxjs-kkd1qa?file=index.ts
– martin
Nov 26 '18 at 9:31
add a comment |
That's because forkJoin requires all source Observables to emit at least one item and when there are no source Observables there's nothing to emit. However, forkJoin will still send the complete notification so you can use for example defaultIfEmpty operator to make sure it always emits at least one next.
forkJoin(observables).pipe(
defaultIfEmpty(null),
).subscribe(...);
Demo: https://stackblitz.com/edit/rxjs-kkd1qa?file=index.ts
answered Nov 26 '18 at 9:19
martinmartin
46.5k1193137
That's because forkJoin requires all source Observables to emit at least one item and when there are no source Observables there's nothing to emit. However, forkJoin will still send the complete notification so you can use for example defaultIfEmpty operator to make sure it always emits at least one next.
forkJoin(observables).pipe(
defaultIfEmpty(null),
).subscribe(...);
Demo: https://stackblitz.com/edit/rxjs-kkd1qa?file=index.ts
answered Nov 26 '18 at 9:19
martinmartin
46.5k1193137
edited Nov 26 '18 at 9:31
answered Nov 26 '18 at 9:19
martinmartin
46.5k1193137
answered Nov 26 '18 at 9:19
martinmartin
46.5k1193137
answered Nov 26 '18 at 9:19
martinmartin
46.5k1193137
46.5k1193137
Thanks Martin. Is the "will still send the complete notification" related only to theforkJoinoperator or is it more general?
– LppEdd
Nov 26 '18 at 9:24
That's related toforkJoin's internal functionality.
– martin
Nov 26 '18 at 9:24
I asked because I was looking at this page learnrxjs.io/operators/combination/forkjoin.html and found nothing mentioning this. Oh btw, had to usedefaultIfEmpty()
– LppEdd
Nov 26 '18 at 9:26
1
This is as edge case so I'm not surprised it's not mentioned there. I don't think this behavior is mentioned not even in the official documentation. You can see that it really works here stackblitz.com/edit/rxjs-kkd1qa?file=index.ts
– martin
Nov 26 '18 at 9:31
add a comment |
Thanks Martin. Is the "will still send the complete notification" related only to theforkJoinoperator or is it more general?
– LppEdd
Nov 26 '18 at 9:24
That's related toforkJoin's internal functionality.
– martin
Nov 26 '18 at 9:24
I asked because I was looking at this page learnrxjs.io/operators/combination/forkjoin.html and found nothing mentioning this. Oh btw, had to usedefaultIfEmpty()
– LppEdd
Nov 26 '18 at 9:26
1
This is as edge case so I'm not surprised it's not mentioned there. I don't think this behavior is mentioned not even in the official documentation. You can see that it really works here stackblitz.com/edit/rxjs-kkd1qa?file=index.ts
– martin
Nov 26 '18 at 9:31
Thanks Martin. Is the "will still send the complete notification" related only to the
forkJoin operator or is it more general?– LppEdd
Nov 26 '18 at 9:24
Thanks Martin. Is the "will still send the complete notification" related only to the
forkJoin operator or is it more general?– LppEdd
Nov 26 '18 at 9:24
That's related to
forkJoin's internal functionality.– martin
Nov 26 '18 at 9:24
That's related to
forkJoin's internal functionality.– martin
Nov 26 '18 at 9:24
I asked because I was looking at this page learnrxjs.io/operators/combination/forkjoin.html and found nothing mentioning this. Oh btw, had to use
defaultIfEmpty()– LppEdd
Nov 26 '18 at 9:26
I asked because I was looking at this page learnrxjs.io/operators/combination/forkjoin.html and found nothing mentioning this. Oh btw, had to use
defaultIfEmpty()– LppEdd
Nov 26 '18 at 9:26
1
1
This is as edge case so I'm not surprised it's not mentioned there. I don't think this behavior is mentioned not even in the official documentation. You can see that it really works here stackblitz.com/edit/rxjs-kkd1qa?file=index.ts
– martin
Nov 26 '18 at 9:31
This is as edge case so I'm not surprised it's not mentioned there. I don't think this behavior is mentioned not even in the official documentation. You can see that it really works here stackblitz.com/edit/rxjs-kkd1qa?file=index.ts
– martin
Nov 26 '18 at 9:31
add a comment |
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