Rx java 如何在RxJava2中无提示地跳过异常？

我有这样一个数据流：

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()

 Observable.fromFuture(it).onErrorResumeNext(Observable.empty())

但如果该“死”值出现在流中，则失败：

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)
...

我是RX的新手，所以很快就搜索到了解决方案：

OneExceptionResumeNext

：

Observable.fromFuture（it）

-->

Observable.fromFuture（it）.OneExceptionResumeNext{Observable.empty（）}

。但是现在我的应用程序永远挂起（在生成我期望的输出之后）。 看起来这条河永远不会结束

我应该以某种方式“关闭”可观察到的

吗？
或者，更一般地说，在使用RX时，这是一种好方法吗？
我应该用另一种方式重新思考吗？
像这样接受例外：

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()

 Observable.fromFuture(it).onErrorResumeNext(Observable.empty())

以下是一些例外情况：

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()

 Observable.fromFuture(it).onErrorResumeNext(Observable.empty())

这不会真的打断这条链条并取消对可观察者的订阅吗？@FabioA。
Future
已返回，但有一个错误。没有别的事可做了。如果您将其用于反应流，则错误是终端事件，链已经停止。@TassosBassoukos，我误解了这个问题。我认为OP希望忽略错误，但继续处理。我弄错了，谢谢你的回复。这会不会真的打断这条链条并取消订阅《观察者》@FabioA。
Future
已返回，但有一个错误。没有别的事可做了。如果您将其用于反应流，则错误是终端事件，链已经停止。@TassosBassoukos，我误解了这个问题。我认为OP希望忽略错误，但继续处理。我错了，谢谢你的回复。