Question

我使用rxmqtt（使用rxjava和paho）与mqtt代理进行通信。我使用javax接受休息请求并将一些内容发布到代理并等待响应。如果我一次发出一个请求，下面的代码工作正常，但如果我有多个并发请求，它只返回最后一个请求的响应，其他请求属于超时异常。

mqttConn.getMqttMessages（）返回一个已经订阅了我需要的所有主题的flowable：

public Flowable<MqttMessage> getMqttMessages() {
    return this.obsClient.subscribe("pahoRx/fa/#", 1);
}

和MqttConnection是一个单例，因为我只需要一个连接到代理，所有发布都在这个连接中完成我注意到我的queryParam id在Web服务请求的每个线程执行中都是不同的（预期的行为），但是当它进入代码的订阅部分时，它只考虑最后一个id值并且没有通过我的验证takeUntil方法：

mqttConn.getMqttMessages().timeout(20, TimeUnit.SECONDS).takeUntil(msgRcv -> {
        System.out.println("received: " + new String(msgRcv.getPayload()) + " ID: " + id);
        return id.equals(new String(msgRcv.getPayload()));
    }).blockingSubscribe(msgRcv -> {
        final byte[] body = msgRcv.getPayload();
        System.out.println(new String(body)); //printing... but not sending the reponse
        response.set("Message Receiced: " + new String(msgRcv.getPayload()));
        return;

    }, e -> {
        if (e instanceof TimeoutException) {
            response.set("Timeout Occured");

        } else {
            response.set("Some kind of error occured " + e.getLocalizedMessage());
        }
    });

问题是，为什么只考虑每个请求应该拥有自己的独立线程时收到的最后一个id？我已经尝试将mqttConn.getMqttConnection（）作为ThreadLocal对象...没有修复。

完整的WS代码：

@Path("/test")
@GET
public String test(@QueryParam("id") String id) throws InterruptedException, MqttException {

    String funcExec = "pahoRx/fe/";
    String content = "unlock with single connection to broker";
    int qos = 1;
    AtomicReference<String> response = new AtomicReference<String>();

    response.set("Initial Value");
    MqttConnection mqttConn = MqttConnection.getMqttConnection();
    ObservableMqttClient obsClient = mqttConn.getBrokerClient();

    MqttMessage msg = MqttMessage.create(78, content.getBytes(), qos, false);
    String topicPub = funcExec + id;

    obsClient.publish(topicPub, msg).subscribe(t -> {
        System.out.println("Message Published");
    }, e -> {
        System.out.println("Failed to publish message: " + e.getLocalizedMessage());
    });

    mqttConn.getMqttMessages().timeout(20, TimeUnit.SECONDS).takeUntil(msgRcv -> {
        System.out.println("received: " + new String(msgRcv.getPayload()) + " ID: " + id);
        return id.equals(new String(msgRcv.getPayload()));
    }).blockingSubscribe(msgRcv -> {
        final byte[] body = msgRcv.getPayload();
        System.out.println(new String(body)); //printing... but not sending the reponse
        response.set("Message Receiced: " + new String(msgRcv.getPayload()));
        return;

    }, e -> {
        if (e instanceof TimeoutException) {
            response.set("Timeout Occured");

        } else {
            response.set("Some kind of error occured " + e.getLocalizedMessage());
        }
    });
    return response.get();
}

我希望解释清楚！

事先提前

为什么线程使用相同的变量值？ RxJava Mqtt

0 个答案: