CODEWITHSUNDEEP
javaandroidrx-javaobservable
Pavel Synek
Pavel Synek

Reputation: 1229

Chaining Observables without significant return values

I am happily using Retrofit with RxJava in my app, but I don't know how to properly chain network calls which don't really return anything except telling that the call succeeded.

One example is function which sends all updated items to server. First I want to call loginManager.refreshLogin() and then synchronizeInventories() and synchronizeItems(), wait until they are both done and return something indicating the UI can be updated.

Right now, the function looks like this:

public Observable<List<Object>> synchronize() {
    return loginManager.refreshLogin()
                    .flatMap(o -> Observable.merge(
                            synchronizeInventories(),
                            synchronizeItems()
                    ))
                    .toList();
}

refreshLogin() returns Observable.just(new Object()), synchronizeInventories() and synchronizeItems() return Observable.empty().

Everything works now, but I don't think it's the right way to do it.

Upvotes: 3

Views: 995

Answers (2)

rompetroll
rompetroll

Reputation: 4799

You can simply use concat to execute a number of Observables in series.

    final Observable flowObs = Observable.concat(obs1, obs2, obs3);
    flowObs.subscribe(); //will first execute the subscriber function on obs1, then on obs2, then on obs3

Upvotes: 0

Diolor
Diolor

Reputation: 13450

Observable<List<Object>> is returned because synchronizeInventories() and synchronizeItems() return different types of objects thus .merge() returns an Observable<Object>.

If both methods returned same objects then it you be a specific class e.g.:

.flatMap(o -> Observable.merge(Observable.just("a"), Observable.just("b")))
.doOnNext(new Action1<String>() {
    @Override
    public void call(String s) {

    }
 })

A way to avoid the general Object class is to use .zip() and use the helping Pair<> class:

.flatMap(o -> Observable.zip(just("a"), just(123), (s, s2) -> new Pair<>(s, s2)))
.doOnNext(pair -> {
    pair.first // get the first object
    pair.second // geth the second
});

Upvotes: 1

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