Question

我有这样的数据流：

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()

我用Future替换了业务逻辑（实际上返回CompletableFuture.supplyAsync s）。而且，是的，这是Kotlin，但我猜你有意图。

当我评论“死”值（57005，0xDEAD）时，输出为：

但如果流中出现“死”值，则会失败：

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，所以很快就用谷歌搜索解决方案：onExceptionResumeNext：Observable.fromFuture(it) - ＆gt; Observable.fromFuture(it).onExceptionResumeNext { Observable.empty<Int>() }。但现在我的应用程序永远挂起（在产生我期望的输出之后）。看起来流不会结束。

我应该“关闭”Observable以某种方式或什么？或者，更一般地说，使用RX时这是一个好方法吗？我应该以另一种方式重新考虑吗？

Answer 1

吞下这样的例外：

{7C945F97-A2CC-4EEB-87AC-3D23D6C490A0};{A9B1787A-32E4-4D51-819D-B3F879574BCE}

如何在RxJava2中以静默方式跳过异常？

1 个答案: