Mono<Mono<Object>>如何订阅

Question

我试图创建以下链：

Mono<Object1> oneMono = Mono.just("1");
Mono<Object2> twoMono = Mono.just("2");

Mono<Mono<Object5>> resultMono = Mono.zip(oneMono, twoMono, (one, two) -> { 

    Mono<Object3> threeMono = Mono.just("3");
    Mono<Object4> fourMono = Mono.just("4");

    return Mono.zip(threeMono, fourMono, (three, four) -> {
        return "5";
    }
}) // get just Mono<Object5> here?

resultMono.subscribe(mono -> { 
    mono.subscribe();  // ugly double subscribe() !!
});

因此，我需要得到5，但在执行1，2和3，4之前。

代码工作正常，但我想出去。

来自Mono<Mono<Object5>>和double subscribe

致Mono<Object5>和single subscribe。

有适合Zip的模拟flatMap吗？

Answer 1

你已经有了。您只需使用flatMap：

Mono<String> oneMono = Mono.just("1");
Mono<String> twoMono = Mono.just("2");

Mono.zip(oneMono, twoMono, (one, two) -> {
    Mono<String> threeMono = Mono.just("3");
    Mono<String> fourMono = Mono.just("4");

    return Mono.zip(threeMono, fourMono, (three, four) -> {
        return "5";
    });
})
.flatMap(stringMono -> stringMono)
.doOnNext(System.out::println)
.subscribe();