Question

我要在一个Flux后面附加一个flatMap，但是如果我添加其他flatMap，则只会返回最后一个。

// Here is an example of the Mono function

private Mono<MyType> appendFirstMono(Group group) {
    return Mono.just(group)
    .map(MyType::new)
        .flatMap(g -> functionZ(group)
        .map(g::setField));
} 


//This works as expected

public Flux<MyType> function1() {

    return returnData(id)
            .thenMany(service.getData(id))
            .flatMap(this::appendFirstMono);
}

//This does not and only returns the last mono (3rd)

public Flux<MyType> function1() {

    return returnData(id)
            .thenMany(service.getData(id))
            .flatMap(this::appendFirstMono)
            .flatMap(this::appendSecondMono)
            .flatMap(this::appendThirdMono);
}

//I've attempted to fix with this... Doesn't work as expected.   

public Flux<MyType> function1() {

    return returnData(id)
            .thenMany(service.getData(id))
            .flatMap(x -> {
                return Flux.merge(
                    appendFirstMono(x),
                    appendSecondMono(x),
                    appendThirdMono(x)
                );
            });
}

我需要在通量上处理每个Mono函数，但似乎无法使每个函数正常执行并返回。

Answer 1

您可以尝试concat一步一步地处理我的示例

        Flux.concat(getMono(0),getMono(1),getMono(2))
            .map(integer -> {
                System.out.println(integer);
                return integer;
            })
            .subscribe();

}

private Mono<Integer> getMono(Integer a) {
    return Mono.just(a)
            ;
}

这将打印0,1,2

将多个Monos组合成一个助焊剂

1 个答案: