在RxJava2中正确使用Single与控制流
[英]Correct use of Single with control flow in RxJava2
问题描述
I have the following Rx chain in my code: 
我的代码中包含以下Rx链：
public Single<List<Items>> dispatchStuff(){
return mRepository.getList()
.filter(list -> list.size() != 0)
.flatMapSingle(mRepository::operateOnList())
.doOnSuccess(mRepository::deleteList())
.subscribeOn(mSchedulerProvider.io())
.observeOn(mSchedulerProvider.ui());
}
However, I'm not sure I'm using the filtering correctly here. 但是，我不确定我在这里是否正确使用了过滤。 The getList() method will always return a value or error, therefore I think using Single is acceptable. getList()方法将始终返回值或错误，因此我认为使用Single是可以接受的。 However, then I want to perform other operations only if there are elements in the list returned. 但是，只有在返回的列表中有元素时，我才想执行其他操作。 If I filter Single and my list size is 0, it will through an exception: 如果我过滤了Single并且列表大小为0，则会出现异常：
java.util.NoSuchElementException
at io.reactivex.internal.operators.maybe.MaybeFlatMapSingle$FlatMapMaybeObserver.onComplete(MaybeFlatMapSingle.java:106)
at io.reactivex.internal.operators.maybe.MaybeFilterSingle$FilterMaybeObserver.onSuccess(MaybeFilterSingle.java:92)
at io.reactivex.internal.operators.single.SingleFromCallable.subscribeActual(SingleFromCallable.java:37)
at io.reactivex.Single.subscribe(Single.java:2692)
at io.reactivex.internal.operators.maybe.MaybeFilterSingle.subscribeActual(MaybeFilterSingle.java:40)
at io.reactivex.Maybe.subscribe(Maybe.java:3707)
at io.reactivex.internal.operators.maybe.MaybeFlatMapSingle.subscribeActual(MaybeFlatMapSingle.java:47)
at io.reactivex.Single.subscribe(Single.java:2692)
at io.reactivex.internal.operators.single.SingleDoOnSuccess.subscribeActual(SingleDoOnSuccess.java:35)
at io.reactivex.Single.subscribe(Single.java:2692)
at io.reactivex.internal.operators.single.SingleSubscribeOn$SubscribeOnObserver.run(SingleSubscribeOn.java:89)
at io.reactivex.Scheduler$1.run(Scheduler.java:138)
at io.reactivex.internal.schedulers.ScheduledRunnable.run(ScheduledRunnable.java:59)
at io.reactivex.internal.schedulers.ScheduledRunnable.call(ScheduledRunnable.java:51)
at java.util.concurrent.FutureTask.run(FutureTask.java:237)
at java.util.concurrent.ScheduledThreadPoolExecutor$ScheduledFutureTask.run(ScheduledThreadPoolExecutor.java:269)
at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1113)
at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:588)
at java.lang.Thread.run(Thread.java:818)
How should I correctly implement this control flow when I'm working with Single ? 使用Single时，应如何正确实现此控制流程？
1 个解决方案
解决方案1
2 已采纳 2017-06-09 18:02:06
Well, it's no longer a Single , it's a Maybe . 好吧，它不再是Single ，而是Maybe 。 If however you don't want to switch to Maybe, perhaps this will help: 但是，如果您不想切换到Maybe，则可能会有所帮助：
public Single<List<Items>> dispatchStuff(){
return mRepository.getList()
.flatMapSingle(list ->
list.size() > 0
? mRepository.operateOnList(list)
.doOnSuccess(mRepository::deleteList)
: Single.just(list)
)
.subscribeOn(mSchedulerProvider.io())
.observeOn(mSchedulerProvider.ui());
}
However, perhaps instead of a Single<List<Items>> you should actually have a Flowable<Items> ? 但是，实际上应该代替Flowable<Items>而不是Single<List<Items>>吗？
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