如果RxJava observable需要很长时间，你如何显示微调器？

Question

I feel like someone has to have tried this, but I can't figure out a nice way to do something if an observable takes to long.

This is the flow I want.

Start a search.
If the search takes longer than some time,
    show a spinner or show progress bar.
When finished do subscription action and hide spinner/progress bar.

The closest I can think of is like a Zip

manager.search(searchTerm)
       .zip(Observable.Timer(1, TimeUnit.SECONDS))
       .subscribe(
           // if the search is non null then we are good
           // if the long time is non 0 we need to show spinner
       );

Is there something better to do? I have been trying all day with no success. In a perfect world I feel like I would want something like

manager.search(searchTerm)
       .timeout(i -> /* do timeout stuff */, 1, TimeUnit.SECONDS)
       .subscribe(item -> /* do search result stuff */);

Answer 1

You can do this by publishing the search Observable through the timeout:

Observable<Integer> source = Observable.just(1).delay(5, TimeUnit.SECONDS);

source
.doOnSubscribe(() -> System.out.println("Starting"))
.publish(o -> 
    o.timeout(1, TimeUnit.SECONDS, Observable.<Integer>fromCallable(() -> {
        System.out.println("Spinning...");
        return null;
    })).ignoreElements().mergeWith(o)
)
.toBlocking()
.subscribe(v -> {
    System.out.println("Hide spinner if shown.");
    System.out.println(v);
});

This works by splitting the source into two hot lanes: the first lane will run a timeout operator which when times out, starts another Observable with the side-effect that shows the spinning control. One of the ways is to use fromCallable for this and ignore its result (this also avoid duplication). The second lane will be unchanged and merged with the timeout lane to deliver the actual value.

Answer 2

Today i found a bit odd but working solution. Idea is to use interval instead of timer.

    fun <T> base_delayed_progress_observable(source: Observable<T>): Observable<T>
    {
        val timer = Observable.interval(100, TimeUnit.MILLISECONDS) //Creates observable that will emit Long++ each 100 miliseconds
                .subscribeOn(Schedulers.io())
                .observeOn(AndroidSchedulers.mainThread())
                .doOnNext(
                    {
                        if (it == 10L)//Here we check current timer value. For example here i check if it is 1 second gone (100 miliseconds * 10 = 1 second)
                        {
                            //here we put all we need to show progress/spinner an so on
                        }
                    })

        return Observable.zip(source, timer,
            BiFunction<T, Long, T> { t1, t2 ->
                //Here we return our original Obervable zipped with timer
                //Timer will be cancelled when our source Observable gets to OnComplete
                return@BiFunction t1
            }).doFinally(
            {
                //Here we can dismiss all progress dilogs/spinner
            })
    }