Question

 List<Integer> list = new ArrayList<Integer>();
    for(int j=1;j<=3;j++)
        list.add(j);


    Observable<Integer> observable = Observable.fromIterable(list)
    .replay()
    .autoConnect();



    observable.subscribe(new Consumer<Integer>() {
        @Override
        public void accept(Integer integer) throws Exception {
            Log.v("consumer1:", ""+integer);
        }
    });


    observable.subscribe(new Consumer<Integer>() {
@Override
public void accept(Integer integer) throws Exception {
    Log.v("consumer2:", ""+integer);

}
});



    observable.subscribe(new Consumer<Integer>() {
        @Override
        public void accept(Integer integer) throws Exception {
            Log.v("consumer3:", ""+integer);
        }
    });

当我运行上面的代码时，我得到以下输出：

consumer1:: 1
consumer1:: 2
consumer1:: 3
consumer2:: 1
consumer2:: 2
consumer2:: 3
consumer3:: 1
consumer3:: 2
consumer3:: 3

我期望重播实际上是“重播”以前发生的所有历史记录。所以我期望发出所有以前发生的流。特别是这是我期望的输出：

//first time nothing to replay so just do the work
consumer1:: 1
consumer1:: 2
consumer1:: 3

//replay consumer1 stream:,
consumer2:: 1
consumer2:: 2
consumer2:: 3
//already replayed now do the work
consumer2:: 1
consumer2:: 2
consumer2:: 3

//replay consumer1 stream:
consumer3:: 1
consumer3:: 2
consumer3:: 3

//replay consumer2 stream:
consumer3:: 1
consumer3:: 2
consumer3:: 3
//now do the work
consumer3:: 1
consumer3:: 2
consumer3:: 3

这是我期望重放发生的事情。我究竟做错了什么？它像它甚至没有以我现在的方式工作。

Answer 1

不确定为什么要执行这种奇特的行为，但是您可以重复使用的数字越来越多：

List<Integer> list = new ArrayList<Integer>();
for (int j = 1; j <= 3; j++) {
    list.add(j);
}


AtomicInteger count = new AtomicInteger();

Observable<Integer> observable = 
    Observable.defer(() -> {
        Observable.fromIterable(list)
        .replay()
        .autoConnect()
        .repeat(count.incrementAndGet());
    });

observable.subscribe(new Consumer<Integer>() {
    @Override
    public void accept(Integer integer) throws Exception {
        Log.v("consumer1:", ""+integer);
    }
});

observable.subscribe(new Consumer<Integer>() {
    @Override
    public void accept(Integer integer) throws Exception {
        Log.v("consumer2:", "" + integer);
    }
});

observable.subscribe(new Consumer<Integer>() {
    @Override
    public void accept(Integer integer) throws Exception {
        Log.v("consumer3:", ""+integer);
    }
});

RxJava2可连接可观察-重放不发出所有先前的项目

1 个答案: