你能不能压缩一个单声道和一个磁通，并为每个磁通值重复这个单声道的值？

是否可以执行类似以下代码的操作？我有一个进行API调用的服务和另一个返回值流的服务。我需要根据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
                     });

我尝试对Mono使用.Repeat()，但它在每次有新的Flux值时都会调用该方法，而且它是一个API调用，所以它不是很好。

可以吗？

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解决方案

这将说明如何将助焊剂与单声道组合在一起，以便每次助焊剂发射时，单声道也会发射。

假设您有一个通量和一个单声道，如下所示：

 // 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，我将向您展示我尝试过的压缩2的错误方法，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();