How to await when all Disposable elements will be finished?

Question

Lets consider following code:

List<Mono<String>> monoList= apiCall();
List<Disposable> disposableList = monoList.stream()
        .map(m-> m.subscribe(str-> {
                           log.info("Mono is finished with "+ str);
                        })
        ).collect(Collectors.toList());
// I need to await here

I need to await when all mono will be finished.

How could I achieve it?

score 1 · Answer 1 · answered Nov 18 '19 at 12:44

1

Not mixing different streaming APIs you could utilize side effects instead of subscriptions and await completion with then()

Mono<Void> await = Flux
            .fromIterable(apiCall())
            .flatMap(m -> m.doOnSuccess(
                str -> log.info("Mono is finished with "+ str)
            )).then()

answered Nov 18 '19 at 12:44

tynn

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An error would result in `await` to complete. If you want to wait for all to terminate, you might want to look into `flatMapDelayError`. – tynn Nov 18 '19 at 13:06
**An error would result in await to complete** could you please rephrase ? – gstackoverflow Nov 19 '19 at 11:37
An error in a single `Mono` would terminate the final `Mono` with an error. – tynn Nov 19 '19 at 11:42

How to await when all Disposable elements will be finished?

1 Answers1