RxJava parallel emit once











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i try to get an array of flowable get executed in parallel with FlowableFromArray



and emit the FlowableFromArray when all flowables are done.



But I'm missing the method to emit on last/latest.



I could only manage to make it work to emit onNext @see below



    val results = FlowableFromArray(flowableArray).parallel()
.runOn(Schedulers.io())
.sequential()
.zipWith(r, BiFunction { t1: Flowable<String>, t2: Int
->
t1
.subscribeOn(Schedulers.io())
.map { i -> parseYoutubeTrack(i) }
.observeOn(AndroidSchedulers.mainThread())
.subscribe { next -> TLog.i(TAG, "_NEXT_ ${next.videoId}") }

})
.subscribe()









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  • This makes no sense; you go parallel to then go back to sequential, then you subscribe in the zip handler. And also what is r?
    – akarnokd
    Nov 11 at 10:32










  • Made it work like so return FlowableFromArray(flowableArray) .parallel() .runOn(Schedulers.computation()) .map { f -> parseYoutubeTrack(f.blockingFirst()) } .sequential() .blockingLatest().toMutableList()
    – Plagueis
    Nov 18 at 17:05

















up vote
0
down vote

favorite












i try to get an array of flowable get executed in parallel with FlowableFromArray



and emit the FlowableFromArray when all flowables are done.



But I'm missing the method to emit on last/latest.



I could only manage to make it work to emit onNext @see below



    val results = FlowableFromArray(flowableArray).parallel()
.runOn(Schedulers.io())
.sequential()
.zipWith(r, BiFunction { t1: Flowable<String>, t2: Int
->
t1
.subscribeOn(Schedulers.io())
.map { i -> parseYoutubeTrack(i) }
.observeOn(AndroidSchedulers.mainThread())
.subscribe { next -> TLog.i(TAG, "_NEXT_ ${next.videoId}") }

})
.subscribe()









share|improve this question






















  • This makes no sense; you go parallel to then go back to sequential, then you subscribe in the zip handler. And also what is r?
    – akarnokd
    Nov 11 at 10:32










  • Made it work like so return FlowableFromArray(flowableArray) .parallel() .runOn(Schedulers.computation()) .map { f -> parseYoutubeTrack(f.blockingFirst()) } .sequential() .blockingLatest().toMutableList()
    – Plagueis
    Nov 18 at 17:05















up vote
0
down vote

favorite









up vote
0
down vote

favorite











i try to get an array of flowable get executed in parallel with FlowableFromArray



and emit the FlowableFromArray when all flowables are done.



But I'm missing the method to emit on last/latest.



I could only manage to make it work to emit onNext @see below



    val results = FlowableFromArray(flowableArray).parallel()
.runOn(Schedulers.io())
.sequential()
.zipWith(r, BiFunction { t1: Flowable<String>, t2: Int
->
t1
.subscribeOn(Schedulers.io())
.map { i -> parseYoutubeTrack(i) }
.observeOn(AndroidSchedulers.mainThread())
.subscribe { next -> TLog.i(TAG, "_NEXT_ ${next.videoId}") }

})
.subscribe()









share|improve this question













i try to get an array of flowable get executed in parallel with FlowableFromArray



and emit the FlowableFromArray when all flowables are done.



But I'm missing the method to emit on last/latest.



I could only manage to make it work to emit onNext @see below



    val results = FlowableFromArray(flowableArray).parallel()
.runOn(Schedulers.io())
.sequential()
.zipWith(r, BiFunction { t1: Flowable<String>, t2: Int
->
t1
.subscribeOn(Schedulers.io())
.map { i -> parseYoutubeTrack(i) }
.observeOn(AndroidSchedulers.mainThread())
.subscribe { next -> TLog.i(TAG, "_NEXT_ ${next.videoId}") }

})
.subscribe()






android rx-java rx-java2






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asked Nov 10 at 21:07









Plagueis

63




63












  • This makes no sense; you go parallel to then go back to sequential, then you subscribe in the zip handler. And also what is r?
    – akarnokd
    Nov 11 at 10:32










  • Made it work like so return FlowableFromArray(flowableArray) .parallel() .runOn(Schedulers.computation()) .map { f -> parseYoutubeTrack(f.blockingFirst()) } .sequential() .blockingLatest().toMutableList()
    – Plagueis
    Nov 18 at 17:05




















  • This makes no sense; you go parallel to then go back to sequential, then you subscribe in the zip handler. And also what is r?
    – akarnokd
    Nov 11 at 10:32










  • Made it work like so return FlowableFromArray(flowableArray) .parallel() .runOn(Schedulers.computation()) .map { f -> parseYoutubeTrack(f.blockingFirst()) } .sequential() .blockingLatest().toMutableList()
    – Plagueis
    Nov 18 at 17:05


















This makes no sense; you go parallel to then go back to sequential, then you subscribe in the zip handler. And also what is r?
– akarnokd
Nov 11 at 10:32




This makes no sense; you go parallel to then go back to sequential, then you subscribe in the zip handler. And also what is r?
– akarnokd
Nov 11 at 10:32












Made it work like so return FlowableFromArray(flowableArray) .parallel() .runOn(Schedulers.computation()) .map { f -> parseYoutubeTrack(f.blockingFirst()) } .sequential() .blockingLatest().toMutableList()
– Plagueis
Nov 18 at 17:05






Made it work like so return FlowableFromArray(flowableArray) .parallel() .runOn(Schedulers.computation()) .map { f -> parseYoutubeTrack(f.blockingFirst()) } .sequential() .blockingLatest().toMutableList()
– Plagueis
Nov 18 at 17:05



















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