Get the latest value of an Observable and emit it

问题:

I'm trying to get the latest value of a given Observable and get it to emit immediately once it's called. Given the code below as an example:

return Observable.just(myObservable.last())
    .flatMap(myObservable1 -> {
        return myObservable1;
    })
    .map(o -> o.x) // Here I want to end up with a T object instead of Observable<T> object

This does not work because by doing this the flatMap will emit myObservable1 which in turn will have to emit to reach the map. I don't know if doing such thing is even possible. Do anyone have any clue on how to achieve this goal? Thank you

回答1:

last() method will not be of any help here as it waits for the Observable to terminate to give you the last item emitted.

Assuming that you do not have the control over the emitting observable you could simply create a BehaviorSubject and subscribe it to the observable that emits the data that you want to listen and then subscribe to the created subject. Since Subject is both Observable and Subscriber you will get what you want.

I think (do not have the time to check it now) you may have to manually unsubscribe from the original observable as the BehaviorSubject once all of his subscribers unsubscribe will not unsubscribe automatically.

Something like this:

BehaviorSubject subject = new BehaviorSubject();
hotObservable.subscribe(subject);
subject.subscribe(thing -> {
    // Here just after subscribing 
    // you will receive the last emitted item, if there was any.
    // You can also always supply the first item to the behavior subject
});

http://reactivex.io/RxJava/javadoc/rx/subjects/BehaviorSubject.html

回答2:

In RxJava, subscriber.onXXX is called asynchronous.It means that if your Observable emit items in new thread, you can never get the last item before return, except you block the thread and wait for the item.But if the Observable emit item synchronously and you dont' change it's thread by subscribeOn and observOn, such as the code:

Observable.just(1,2,3).subscribe();

In this case, you can get the last item by doing like this:

Integer getLast(Observable<Integer> o){
    final int[] ret = new int[1];
    Observable.last().subscribe(i -> ret[0] = i);
    return ret[0];
}

It's a bad idea doing like this.RxJava prefer you to do asynchronous work by it.

回答3:

What you actually want to achieve here is to take an asynchronous task and transform it to a synchronous one.

There are several ways to achieve it, each one with it's pros and cons:

• Use toBlocking() - it means that this thread will be BLOCKED, until the stream is finish, in order to get only one item simply use first() as it will complete once an item is delivered. let's say your entire stream is Observable<T> getData(); then a method that will get the last value immediately will look like this:

public T getLastItem(){ return getData().toBlocking().first(); }

please don't use last() as it will wait for the stream to complete and only then will emit the last item.

If your stream is a network request and it didn't get any item yet this will block your thread!, so only use it when you are sure that there is an item available immediately (or if you really want a block...)

• another option is to simply cache the last result, something like this:

  getData().subscribe(t-> cachedT = t;) //somewhere in the code and it will keep saving the last item delivered public T getLastItem(){ return cachedT; }

if there wasn't any item sent by the time you request it you will get null or whatever initial value you have set. the problem with this approch is that the subscribe phase might happen after the get and might make a race condition if used in 2 different threads.