急切地评估。switchifempty（）有什么意义？

Question

即使我的流不是空的，回退流总是会被创建？这样做背后的意图是什么？这是极不地道的。

  public static void main(String[] args) {
    Mono<Integer> m = Mono.just(1);
    m.flatMap(a -> Mono.delay(Duration.ofMillis(5000)).flatMap(p -> Mono.empty()))
        .switchIfEmpty(getFallback())
        .doOnNext(a -> System.out.println(a))
        .block();
  }

  private static Mono<Integer> getFallback() {
    System.out.println("In Here");
    return Mono.just(5);
  }

In Here (printed immediately)
5 (after 5s)

Answer 1

这里您需要理解的是组装时间和订阅时间之间的区别。

组装时间是通过构建运算符链来创建管道的时间。此时，您的发布服务器尚未订阅，您需要进行必要的思考。

订阅时间是指通过订阅触发执行并且数据开始流经管道的时间。这时，您需要从回调、lambdas、延迟执行等方面进行反应性思考。

public static void main(String[] args) {
    Mono<Integer> m = Mono.just(1);

    m.flatMap(a -> Mono.delay(Duration.ofMillis(5000)).flatMap(p -> Mono.empty()))
     .switchIfEmpty(getFallback())
     .doOnNext(a -> System.out.println(a))
     .block();
}

private static Mono<Integer> getFallback() {
    System.out.println("Assembly time, here we are just in the process of creating the mono but not triggering it. This is always called regardless of the emptiness of the parent Mono.");
    return Mono.fromCallable(() -> {
        System.out.println("Subscription time, this is the moment when the publisher got subscribed. It is got called only when the Mono was empty and fallback needed.");
        return 5;
    });
}

2,您可以使用Mono.defer并将内单声道的执行和组装延迟到订阅：

public static void main(String[] args) {
    Mono<Integer> m = Mono.just(1);
    m.flatMap(a -> Mono.delay(Duration.ofMillis(5000)).flatMap(p -> Mono.empty()))
     .switchIfEmpty(Mono.defer(() -> getFallback()))
     .doOnNext(a -> System.out.println(a))
     .block();
}

private static Mono<Integer> getFallback() {
    System.out.println("Since we are using Mono.defer in the above pipeline, this message gets logged at subscription time.");
    return Mono.just(5);
}

请注意，您最初的解决方案也非常好。您只需要知道返回Mono之前的代码是在汇编时执行的。