Can you Flux.zip a mono and a flux and and repeat the mono value for every flux value?

问题

Is it possible to do something like the code below? I have one service that makes an API call and another one which returns a stream of values. I need to modify every value by the value returned by the API call.

return Flux.zip(
                     someMono.get(),
                     someFlux.Get(),
                     (d, t) -> {
                         //HERE D IS ALWAYS THE SAME AND T IS EVERY NEW FLUX VALUE
                     });

I've tried with .repeat() for the Mono and it works, but it's calling the method every time there's a new Flux value and it's a API call, so it's not good.

Is it possible?

回答1:

This will illustrate how to combine a flux with a mono such that every time the flux emits, the mono is emitted as well.

Suppose you have a flux and a mono like this:

 // a flux that contains 6 elements.
 final Flux<Integer> userIds = Flux.fromIterable(List.of(1,2,3,4,5,6));

 // a mono of 1 element.
 final Mono<String> groupLabel = Mono.just("someGroupLabel");

First, I'll show you the wrong way of trying to zip the 2 which I tried, and I think other people would try:

 // wrong way - this will only emit 1 event 
 final Flux<Tuple2<Integer, String>> wrongWayOfZippingFluxToMono = userIds
         .zipWith(groupLabel);

 // you'll see that onNext() is only called once, 
 //     emitting 1 item from the mono and first item from the flux.
 wrongWayOfZippingFluxToMono
         .log()
         .subscribe();

 // this is how to zip up the flux and mono how you'd want, 
 //     such that every time the flux emits, the mono emits. 
 final Flux<Tuple2<Integer, String>> correctWayOfZippingFluxToMono = userIds
         .flatMap(userId -> Mono.just(userId)
                 .zipWith(groupLabel));

 // you'll see that onNext() is called 6 times here, as desired. 
 correctWayOfZippingFluxToMono
         .log()
         .subscribe();

回答2:

You can accomplish this using cache operator. Uncomment the Flux without cache and you will see the number of calls made to getNum == 100. With cache it will be 1.

public class RepeatableMono {

  private static AtomicInteger numberOfCalls = new AtomicInteger(0);

  static Integer getNum() {
    System.out.println("GetNum Called: " + numberOfCalls.incrementAndGet());
    return 0;
  }

  public static void main(String[] args) {
    // This will print `GetNum Called: ` 100 times.
    //Flux<Integer> neverEndingFlux = Mono.defer(() -> Mono.just(getNum()))
    // .repeat();

    // This will print `GetNum Called: ` 1 times.
    Flux<Integer> neverEndingFlux = Mono.defer(() -> Mono.just(getNum()))
        .cache()
        .repeat();

    Flux<Integer> stream = Flux.range(1, 100);

    Flux.zip(neverEndingFlux, stream, (x, y) -> x + " " + y)
        .subscribe(System.out::println);
  }
}

来源：https://stackoverflow.com/questions/59742934/can-you-flux-zip-a-mono-and-a-flux-and-and-repeat-the-mono-value-for-every-flux

标签

java

spring-webflux

spring-reactive