Question

受到T.Nurkiewicz＆＃34;反应性编程与RxJava的启发＆＃34;我试图在我正在进行的项目中应用它，这是我面临的问题。

我有一个Rest端点，它接受输入流和用户名，并返回更新用户名的链接或返回错误请求错误。以下是我尝试使用RxJava实现此目的的方法：

    @PUT
    @Path("{username}")
    public Response updateCredential(@PathParam("username") final String username, InputStream stream) {
        CredentialCandidate candidate = new CredentialCandidate();
        Observable.just(repository.getByUsername(username))
                .subscribe(
                    credential -> {
                            serializeCandidate(candidate, stream);
                            try {
                                repository.updateCredential(build(credential, candidate));
                            } catch (Exception e) {
                                String msg = "Failed to update credential +\""+username+"\": "+e.getMessage();
                                throw new BadRequestException(msg, Response.status(Response.Status.BAD_REQUEST).build());
                            }
                        },
                        ex -> {
                            String msg = "Couldn't update credential \""+username+"\""
                            + ". A credential with such username doesn't exist: " + ex.getMessage();
                            logger.error(msg);
                            throw new BadRequestException(msg, Response.status(Response.Status.BAD_REQUEST).build());
                });//if the Observable completes without exceptions we have a success case
        Map<String, String> map = new HashMap<>();
        map.put("path", "credential/" + username);
        return Response.ok(getJsonRepr("link", uriGenerator.apply(appsUriBuilder, map).toASCIIString())).build();
}

我的问题出在第11行（onNext方法的catch子句）。这是日志输出，可以快速显示发生的情况：

19:23:50.472 [http-listener(4)] ERROR com.vgorcinschi.rimmanew.rest.services.CredentialResourceService             - Couldn't update credential "admin". A credential with such username doesn't exist: Failed to update credential +"admin": Password too weak!

因此，onNext方法中抛出的异常进入上游并以onError方法结束！显然this works as designed，但我对如何返回错误请求错误的正确原因感到困惑。毕竟在我的测试用例中，存储库找到了用户的凭证，正确的错误是建议的密码太弱了。这是生成错误的辅助方法：

private Credential build(Credential credential, CredentialCandidate candidate) {
         if(!isOkPsswd.test(candidate.getPassword())){
            throw new BadRequestException("Password too weak!", Response.status(Response.Status.BAD_REQUEST).build());
        }
...
}

我仍然是反应式编程的新手，所以我意识到我可能会遗漏一些显而易见的东西。浏览这本书并没有让我得到答案，所以我将不胜感激。

以防万一，这是完整的堆栈跟踪：

updateCredentialTest(com.vgorcinschi.rimmanew.services.CredentialResourceServiceTest)  Time elapsed: 0.798 sec  <<< ERROR!
rx.exceptions.OnErrorFailedException: Error occurred when trying to propagate error to Observer.onError
    at com.vgorcinschi.rimmanew.rest.services.CredentialResourceService.lambda$updateCredential$9(CredentialResourceService.java:245)
    at rx.internal.util.ActionSubscriber.onNext(ActionSubscriber.java:39)
    at rx.observers.SafeSubscriber.onNext(SafeSubscriber.java:134)
    at rx.internal.util.ScalarSynchronousObservable$WeakSingleProducer.request(ScalarSynchronousObservable.java:276)
    at rx.Subscriber.setProducer(Subscriber.java:209)
    at rx.Subscriber.setProducer(Subscriber.java:205)
    at rx.internal.util.ScalarSynchronousObservable$JustOnSubscribe.call(ScalarSynchronousObservable.java:138)
    at rx.internal.util.ScalarSynchronousObservable$JustOnSubscribe.call(ScalarSynchronousObservable.java:129)
    at rx.Observable.subscribe(Observable.java:10238)
    at rx.Observable.subscribe(Observable.java:10205)
    at rx.Observable.subscribe(Observable.java:10045)
    at com.vgorcinschi.rimmanew.rest.services.CredentialResourceService.updateCredential(CredentialResourceService.java:238)
    at com.vgorcinschi.rimmanew.services.CredentialResourceServiceTest.updateCredentialTest(CredentialResourceServiceTest.java:140)

Answer 1

似乎你没有正确掌握反应式编程原则。

首先，Observable的API是异步的，而您试图通过尝试直接从方法返回Response值而不是返回{来强制它成为同步API {1}} Observable<Response>通过Response通知随时间发出onNext()值这就是为什么你正在努力解决异常，每个通知lambda方法（onNext / onError）都由Observable机制封装，以便创建一个遵守某些规则的正确流（ Observable contract），其中一些预期的行为是错误应该重定向到onError()方法，这是异常捕获方法，你不应该扔到那里，扔在那里将被视为致命错误并且会被投掷OnErrorFailedException吞下。

理想情况下会是这样的：

public Observable<Response> updateCredential(@PathParam("username") final String username,
                                             InputStream stream) {
    rerurn Observable.fromCallable(() -> {
        CredentialCandidate candidate = new CredentialCandidate();
        Credential credential = repository.getByUsername(username);
        serializeCandidate(candidate, stream);
        repository.updateCredential(build(credential, candidate));
        Map<String, String> map = new HashMap<>();
        map.put("path", "credential/" + username);
        return Response.ok(getJsonRepr("link", uriGenerator.apply(appsUriBuilder, map).toASCIIString())).build();
    })
            .onErrorReturn(throwable -> {
                String msg = "Failed to update credential +\"" + username + "\": " + e.getMessage();
                throw new BadRequestException(msg, Response.status(Response.Status.BAD_REQUEST).build());
            });
}

使用fromCallable以便在订阅时使请求发生（当Observable.just(repository.getByUsername(username))在Observable构造时将同步执行），成功路径与可调用本身有关，而如果发生任何错误，您将使用onErrorReturn运算符将其转换为自定义异常。

通过他的方法，您将返回Observable对象，当您订阅它时，该对象将起作用，您将获得Observable和Reactive方法的所有好处，以便能够将其与其他一些操作组合在一起，能够从外部指定它是同步行动（当前线程）还是异步行动（使用Scheduler）。

关于反应式编程的更详细解释，我建议从AndréStaltz的这个伟大的tutorial开始。

RxJava：OnErrorFailedException。确定正确的原因

1 个答案: