java - 您可以通过 Flux.zip 压缩一个单声道和一个通量，并为每个通量值重复单声道值吗？

Question

是否可以像下面的代码那样做？我有一个进行 API 调用的服务和另一个返回值流的服务。我需要通过 API 调用返回的值修改每个值。

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

我已经为 Mono 尝试使用 .repeat() 并且它有效，但是每次有一个新的 Flux 值时它都会调用该方法并且它是一个 API 调用，所以它不是很好。

这可能吗？

Answer 1

这将说明如何将通量与单声道组合，这样每次通量发射时，单声道也会发射。

假设您有这样的助焊剂和单声道:

 // 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");

首先，我将向您展示尝试压缩 2 的错误方法，我试过，我想其他人也会尝试:

 // 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();