RxJava 的 retryWhen 操作符

问题描述

我正在尝试深入了解 retryWhen 运算符，我有一些代码如下.

I'm trying to understand retryWhen operator in depth and I have some code as below.

    Flowable.just(1, 2, 3, 4, 5)
            .subscribeOn(Schedulers.io())
            .observeOn(AndroidSchedulers.mainThread())
            .retryWhen { throwable ->
                Log.d("Debug", "retryWhen proceed...")
                throw Exception("There is a exception")
            }
            .subscribe(
                    { item ->
                        Log.d("Debug", "success : $item")
                    },
                    { throwable ->
                        Log.d("Debug", "error : ${throwable.message}")
                    },
                    {
                        Log.d("Debug", "complete")
                    }
            )

结果如下所示.

调试:重试继续时...

调试:错误:有异常

问题是什么时候触发retryWhen操作符?

The question is that when retryWhen operator is triggered?

我假设只有在发生异常时才会触发 retryWhen 操作符.

I assume retryWhen operator will be triggered only when there is a exception occurs.

但结果显然不是我想的那样，

But the result is not what I thought obviously,

对此有什么想法吗?谢谢！

Any thoughts on this? Thanks!

推荐答案

retryWhen { errors ->... } 接受一个 Observable 并且应该返回一个 Observable 返回任何重试或错误em>停止重试.

retryWhen { errors -> ... } take an Observable<Throwable> and should return an Observable that return anything for retrying or an error for stop retrying.

一个例子可能是:

.retryWhen(attempts -> {
  return attempts.zipWith(Observable.range(1, 3), (n, i) -> i).flatMap(i -> {
    System.out.println("delay retry by " + i + " second(s)");
    return Observable.timer(i, TimeUnit.SECONDS);
  });
})

(取自 http://reactivex.io/documentation/operators/retry.html)
此代码将延迟每次重试.

(taken from http://reactivex.io/documentation/operators/retry.html)
This code will delay each retry.

顺便说一句，抛出异常不是这个方法要做的事情.

By the way, throwing an exception is not the thing to do in this method.

文档:
* 很棒的博客文章，解释了retryWhen

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