Question

给定以下代码，result.block（）等于＆＃34; xx＆＃34;，isError（）== false，然后运行繁荣错误处理程序，显示堆栈跟踪，未完成承诺。我希望结果是＆＃34; ko＆＃34;。

我做错了什么？ block（）Javadoc说

如果onComplete，
将返回null，如果onNext
，则返回T.

public class RApp {

static final Logger LOG = LoggerFactory.getLogger(RApp.class);

public static void main(String[] args) {

    MonoProcessor<String> texecute = MonoProcessor.create();
    Mono<String> result = texecute.delaySubscription(Duration.ofSeconds(2))
            .onErrorReturn("ko")
            .doOnNext(s -> parse(s)
            .doOnSuccess(p -> LOG.info("promise completed {}", p))
            .doOnTerminate((z, e) -> LOG.info("term value: {} , {}", z, e))
            .doOnError(t -> {
                LOG.error("boom", t);
            })
            .subscribe());

    texecute.onNext("xx");

    LOG.info("...............;");
    String block = result.block();
    LOG.info("r={}", block);
    boolean error = texecute.isError();
    LOG.info(error ? "error" : "no error");
    texecute.dispose();

}

public static Mono<String> parse(String s) {
    System.out.println("parse s = " + s);
    if (s.equals("xx")) {
        return Mono.error(new Exception("no xx"));
    }
    return Mono.just(s);
}
}

Answer 1

回答这个问题：do *是副作用方法，不修改每https://projectreactor.io/docs/core/release/reference/#error.handling的序列，onErrorReturn 的顺序很重要。

下面的正确工作解决方案，带有reactor.core.Exceptions.propagate的奖励以包装已检查的异常和java 8失败计数器：

    LongAdder failureStat = new LongAdder();

    MonoProcessor<String> texecute = MonoProcessor.create();
    Mono<String> result = texecute
            .delaySubscription(Duration.ofSeconds(2))
            .map(e -> parse2(e)).doOnError(e -> {
                failureStat.increment();
            }).doOnSuccess(s -> {
                LOG.info("success {}", s);
            })
            .onErrorReturn("ko")
            .subscribe();

    texecute.onNext("xx");

    LOG.info("...............;");
    String block = result.block();
    LOG.info("r={}", block);
    System.out.println("failureStat = " + failureStat);
    texecute.dispose();

public static String parse2(String s) {
    System.out.println("parse s = " + s);
    if (s.equals("xx")) {
        try {
            throw new Exception("no xx");
        } catch (Exception e) {
            throw Exceptions.propagate(e);
        }
    }
    return s;
}

带有Project Reactor的Java：为什么Mono block（）没有看到错误？

1 个答案: