如何在RxJava2中默默跳过异常?
问题描述:
我有一个这样的数据流:如何在RxJava2中默默跳过异常?
Observable
.fromFuture(
CompletableFuture.supplyAsync { // First remote call returns Future<List<Type>>
listOf(1, 2, 3, 57005, 5)
},
Schedulers.computation()
)
.flatMap { it.toObservable() } // I turn that list into a stream of single values to process them one by one
.map {
CompletableFuture.supplyAsync { // This remote call may fail if it does not like the input. I want to skip that failures and continue the stream like the fail never occurred.
if (it == 0xDEAD) {
throw IOException("Dead value!")
}
it
}
}
.flatMap {
Observable.fromFuture(it) // Turn that Futures into a stream of Observables once again
}
.doOnNext {
println(it) // Debug
}
.blockingSubscribe()
我已经更换了业务逻辑(实际返回Future
S)与CompletableFuture.supplyAsync
。 而且,是的,这是Kotlin,但我想你有这个意图。
当我评论 “死” 值(57005
)输出为:
1
4
9
25
但是,如果 “死” 值出现在流,失败:
1
4
9
Exception in thread "main" java.lang.RuntimeException: java.util.concurrent.ExecutionException: java.io.IOException: Dead value!
at io.reactivex.internal.util.ExceptionHelper.wrapOrThrow(ExceptionHelper.java:45)
at io.reactivex.internal.operators.observable.ObservableBlockingSubscribe.subscribe(ObservableBlockingSubscribe.java:86)
at io.reactivex.Observable.blockingSubscribe(Observable.java:5035)
at by.dev.madhead.rx.TestKt.main(test.kt:41)
Caused by: java.util.concurrent.ExecutionException: java.io.IOException: Dead value!
at java.util.concurrent.CompletableFuture.reportGet(CompletableFuture.java:357)
at java.util.concurrent.CompletableFuture.get(CompletableFuture.java:1895)
...
我一个新手在RX,所以很快搜索出一个解决方案:onExceptionResumeNext
:Observable.fromFuture(it)
- >Observable.fromFuture(it).onExceptionResumeNext { Observable.empty<Int>() }
。但是现在我的应用程序永远挂起(在产生我期望的输出之后)。 看起来像流永远不会结束。
我应该“关机”Observable
莫名其妙或什么? 或者更一般地说,使用RX时是否是一个好方法? 我是否应该以另一种方式重新考虑它?
答
燕子的例外是这样的:
Observable.fromFuture(it).onErrorResumeNext(Observable.empty())
这实际上做的伎俩,愚蠢的我。谢谢! – madhead