Question

我正在尝试创建以下链：

DateTimeOffset

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<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() !! });和Mono<Mono<Object5>>

到double subscribe和Mono<Object5>。

是否存在Zip的模拟flatMap？

Answer 1

嗯，你已经拥有它了。你只需要使用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();

单<单<对象>＆GT;如何订阅

1 个答案: