RxJava 将 Observable 与另一个可选的 Observable 结合使用，并带有超时

Question

Asume we have two observables A and B . A publishes a result certainly, while the result from B might not be published at all (timeout).

The question is how to map the result from A and B if B returns within a timeframe, otherwise just return the result from A .

Observable<DatabaseObject> A = getDatabaseElement();
Observable<NetworkObject> B = restApi.getElement();

Map example:

map((databaseObject, networkObject) => {
  databaseObject.setData(networkObject);
  return databaseObject;
})

Answer 1

In order to timeout B observable use take operator with time argument:

B.take(10, TimeUnit.SECONDS)

In order to receive either A or B (if B is ready within timeout) use concatWith :

A.concatWith(B.take(10, TimeUnit.SECONDS))
    .takeLast(1)

In case you wish to combine A and B (optionally enrich A with B ):

A.concatWith(B.take(10, TimeUnit.SECONDS))
    .reduce((a, b) -> a.setData(b))

In case A and B are of different types (optionally enrich A with B ):

Observable.combineLatest(
    A,
    B.take(10, TimeUnit.SECONDS).defaultIfEmpty(stubB)),
    (a, b) -> { if (b != stubB) a.setData(b); }
)

Answer 2

You might want to take a different approach here .. There is an operator designed exactly for the behavior you described. Take a look at .timeout(long time, TimeUnits units, Observable backupObservable) which instructs a stream to switch from the original observable (in your case B, the network request I assume) to the backup A observable when there are no other items in the original stream for the specified amount of time.