I'm using RabbitMQ (and Celery) on Java, here is my code to get a message from RabbitMQ based on a tutorial I am reading:
QueueingConsumer consumer = new QueueingConsumer(channel);
channel.basicConsume(QUEUE_NAME, true, consumer);
while (true) {
QueueingConsumer.Delivery delivery = consumer.nextDelivery();
String message = new String(delivery.getBody());
System.out.println(" [x] Received '" + message + "'");
}
But I only get a message when the task begins - when I would like to get a message when the task is complete. Any help?
You should not be using QueueingConsumer since it's considered deprecated as explained here: https://www.rabbitmq.com/releases/rabbitmq-java-client/current-javadoc/com/rabbitmq/client/QueueingConsumer.html
On the contrary, you should be creating your own consumer that implements the interface Consumer from RabbitMQ libraries. There is a method you will have to implement called handleDelivery that will be called every time you get a message. Then, to start it you need to call channel.basicConsume(QUEUE_NAME, true, consumer).
Example:
channel.basicConsume(queueName, autoAck, "myConsumerTag", new DefaultConsumer(channel) {
#Override
public void handleDelivery(String consumerTag,
Envelope envelope,
AMQP.BasicProperties properties,
byte[] body) throws IOException
{
//your code here
}
});
Related
I am currently learning how to use RabbitMQ to schedule and dispatch jobs to different VM.I am now working on the worker side. The worker on the VM need to do some hard-loading jobs and return to the server if it is successfully done.
I have done some surveys on the official api and also here ,trying to test if it can work.
Connection connection = factory.newConnection();
final Channel channel = connection.createChannel();
channel.basicQos(10);
Consumer consumer = new DefaultConsumer(channel) {
#Override
public void handleDelivery(String consumerTag, final Envelope envelope, AMQP.BasicProperties properties, final byte[] body) throws IOException {
Thread th = new Thread() {public void run() {
try{
//do some jobs here...
synchronized (this) {channel.basicAck(envelope.getDeliveryTag(), false);}
} catch (Exception e) {
e.printStackTrace();
try {
synchronized (this) {channel.basicReject(envelope.getDeliveryTag(), false)}
} catch (IOException e1) {e1.printStackTrace();}
}
};
th.start();
}
};
channel.basicConsume(queueName, false, consumer);
This code works for me. But I am just wondering if there is a better and thread-safer way to do it.
How about using an ExecutorService instead of a new thread for every message? Depending on the rate of incoming messages the number of threads created by your approach can grow very large very quickly which could bring down your service.
I want a queue to be consumed by only one subscriber at a time. So if one subscriber drops, then another one(s) will have the chance of subscribing.
I am looking for the correct way of doing it in Spring AMQP. I did this in pure Java, based on the example in RabbitMQ's website. I passively declare the queue, check its consumer count, if it is 0, then start to consume it.
Here's the code.
ConnectionFactory factory = new ConnectionFactory();
factory.setHost("localhost");
Connection connection = factory.newConnection();
Channel channel = connection.createChannel();
int count = channel.queueDeclarePassive(QUEUE_NAME).getConsumerCount();
System.out.println("count is "+count);
if (count == 0) {
channel.queueDeclare(QUEUE_NAME, false, false, false, null);
System.out.println(" [*] Waiting for messages. To exit press CTRL+C");
DeliverCallback deliverCallback = (consumerTag, delivery) -> {
String message = new String(delivery.getBody(), StandardCharsets.UTF_8);
System.out.println(" [x] Received '" + message + "'");
};
channel.basicConsume(QUEUE_NAME, true, deliverCallback, consumerTag -> { });
} else{
System.out.println("subscribed by some other processor(s)");
}
I also can check the subscriber count in Spring AMQP this way. But it is too late, because it already listens to the queue.
#RabbitListener(queues = "q1")
public void receivedMessageQ1(String message, Channel channel){
try {
int q1 = channel.queueDeclarePassive("q1").getConsumerCount();
// do something.
} catch (IOException e) {
System.out.println("exception occurred");
}
}
In a nutshell, I want to consume a queue based on its consumer count. I hope I am clear.
Set the exclusive flag on the #RabbitListener; RabbitMQ will only allow one instance to consume. The other instance(s) will attempt to listen every 5 seconds (by default). To increase the interval, set the container factory's recoveryBackOff.
#SpringBootApplication
public class So56319999Application {
public static void main(String[] args) {
SpringApplication.run(So56319999Application.class, args);
}
#RabbitListener(queues = "so56319999", exclusive = true)
public void listen (String in) {
}
#Bean
public Queue queue() {
return new Queue("so56319999");
}
}
I have a JAVA application which creates consumers that listen to rabbitmq . I need to know the started consumer is still working fine and if not then i need to restart the consumer.
Is their any way i can do that. Currently my main application creates an Executor thread pool and passes this executor while creating new connection.
ExecutorService executor = Executors.newFixedThreadPool(30);
Connection connection = factory.newConnection(executor);
The main method then create 30 consumerApp object by calling constructor with new channel as argument and call the listen() method
for(int i=0;i<=30;i++) {
ConsumerApp consumer = new ConsumerApp(i,connection.createChanell());
consumer.listen() }
The listen method in consumerApp listen to a queue and start a DefaultConsumer Object which simply prints the received message
listen() {
try {
channel.queueDeclare("test-queue-name", false, false, false, null);
}
catch {
System.out.println("Exception on creating Queue")
}
Consumer consumer = new DefaultConsumer(this.channel) {
#Override
public void handleDelivery(String consumerTag, Envelope envelope, AMQP.BasicProperties properties,
byte[] body) throws IOException {
String message = new String(body, "UTF-8");
System.out.println(" [x] Received Message in consumer '"+consumerId+" "+ message + "'");
}
};
//Now starting the consumer
try {
channel.basicConsume(QUEUE_NAME, true, consumer);
}
catch (ShutdownSignalException | IOException ex) {
ex.printStackTrace();
}
}
I want to know is their any way i can check the consumer is active . My idea is to catch the shutdown signal exception and recreate the consumer object and recall the listen method . Is this necessary as rabbitmq auto recovers and connnect back. ? But how can i ensure this ?
Is this any way achievable using the threadpool passed to rabbitmq connector.
I am using latest version of rabbitmq client 5.3.0
Consumer has different methods that can help you track the state of your consumer. You're likely to be interested in handleConsumeOk and in handleCancel.
Automatic connection recovery will indeed re-register consumers after a connection failure, but that doesn't prevent you from following their state manually to e.g. expose some information on JMX.
I implementing websockets using Vert.x 3.
The scenario is simple: opening socket from client doing some 'blocking' work at the vertex verticle worker and when finish response with the answer to the client(via the open socket)
Please tell me if I am doing it right:
Created VertxWebsocketServerVerticle. as soon as the websocket is opening and request coming from the client I am using eventBus and passing the message to
EventBusReceiverVerticle. there I am doing blocking operation.
how I am actually sending back the response back to VertxWebsocketServerVerticle and sending it back to the client?
code:
Main class:
public static void main(String[] args) throws InterruptedException {
Vertx vertx = Vertx.vertx();
vertx.deployVerticle(new EventBusReceiverVerticle("R1"),new DeploymentOptions().setWorker(true));
vertx.deployVerticle(new VertxWebsocketServerVerticle());
}
VertxWebsocketServerVerticle:
public class VertxWebsocketServerVerticle extends AbstractVerticle {
public void start() {
vertx.createHttpServer().websocketHandler(webSocketHandler -> {
System.out.println("Connected!");
Buffer buff = Buffer.buffer().appendInt(12).appendString("foo");
webSocketHandler.writeFinalBinaryFrame(buff);
webSocketHandler.handler(buffer -> {
String inputString = buffer.getString(0, buffer.length());
System.out.println("inputString=" + inputString);
vertx.executeBlocking(future -> {
vertx.eventBus().send("anAddress", inputString, event -> System.out.printf("got back from reply"));
future.complete();
}, res -> {
if (res.succeeded()) {
webSocketHandler.writeFinalTextFrame("output=" + inputString + "_result");
}
});
});
}).listen(8080);
}
#Override
public void stop() throws Exception {
super.stop();
}
}
EventBusReceiverVerticle :
public class EventBusReceiverVerticle extends AbstractVerticle {
private String name = null;
public EventBusReceiverVerticle(String name) {
this.name = name;
}
public void start(Future<Void> startFuture) {
vertx.eventBus().consumer("anAddress", message -> {
System.out.println(this.name +
" received message: " +
message.body());
try {
//doing some looong work..
Thread.sleep(10000);
System.out.printf("finished waiting\n");
startFuture.complete();
} catch (InterruptedException e) {
e.printStackTrace();
}
});
}
}
I always get:
WARNING: Message reply handler timed out as no reply was received - it will be removed
github project at: https://github.com/IdanFridman/VertxAndWebSockets
thank you,
ray.
Since you are blocking your websocket handler until it receives a reply for the sent message to the EventBus, which will not, in fact, be received until the set up delay of 10s laps, you certainly will get warning since the reply handler of the event bus will timeout -> Message sent but no response received before the timeout delay.
Actually I don't know if you are just experimenting the Vert.x toolkit or you are trying to fulfill some requirement, but certainly you have to adapt your code to match in the Vert.x spirit:
First you should better not block until a message is received in your websocket handler, keep in mind that everything is asynchrounous when it comes to Vert.x.
In order to sleep for some time, use the Vert.x way and not the Thread.sleep(delay), i.e. vertx.setTimer(...).
I'm testing open MQ for send and receive messages in my project. I have no problem to configure it to send a synchronous message, but i can't find any way in the official documentation to configure the message to be consumed 15 minutes after the producer send a message, and continue call the consumer if an error appears.
offical documentation: http://dlc.sun.com/pdf/819-7757/819-7757.pdf
my method whom send a message
public void sendMessage(EntradaPrecomven entrada){
try{
Hashtable env = new Hashtable();
env.put(Context.INITIAL_CONTEXT_FACTORY, "com.sun.jndi.fscontext.RefFSContextFactory");
env.put(Context.PROVIDER_URL, "file:///C:/mqteste");
// Create the initial context.
Context ctx = new InitialContext(env);
// Look up the connection factory object in the JNDI object store.
autenticisFactory = (ConnectionFactory) ctx.lookup(CF_LOOKUP_NAME);
mdbConn = autenticisFactory.createConnection();
mdbSession = mdbConn.createSession(false, Session.AUTO_ACKNOWLEDGE);
Destination destination = (Destination) ctx.lookup(DEST_LOOKUP_NAME);
MessageProducer myProducer = mdbSession.createProducer(destination);
ObjectMessage outMsg = mdbSession.createObjectMessage(entrada);
outMsg.setJMSRedelivered(Boolean.TRUE);
myProducer.send(outMsg);
consumidor = mdbSession.createConsumer(destination);
MessageMDB myListener = new MessageMDB();
consumidor.setMessageListener(myListener);
mdbConn.start();
mdbConn.close();
}catch(Exception e){
try {
mdbSession.rollback();
} catch (JMSException e1) {}
}
}
My listener:
#Override
public void onMessage(Message msg) {
ObjectMessage objMessage = (ObjectMessage) msg;
try {
System.out.println("Received Phone Call:" + objMessage.getJMSRedelivered());
throw new JMSException("TESTE");
} catch (JMSException e) {
e.printStackTrace();
}
}
So, when i call mdbConn.start() the sendMessage() is called, but i want to call 15 minutes after the call. And whatever it sendMessage() does, the message is always removed from the queue. How can i keep the messagen in queue to be called later ?
Thanks!
The message is removed from the broker queue due to the fact that the session you are using is set to auto acknowledge.
mdbSession = mdbConn.createSession(false, Session.AUTO_ACKNOWLEDGE);
This will automatically send an acknowledgement to the broker that the listener has received a message for the consumer with which it is associated once the onMessage() method has executed to completion. This then results in the message being removed from the queue.
If you manually take over the acknowledgement process you can choose to only acknowledge the receipt of the message at a time of your choosing (be that 15 minutes later or whatever criteria you have for the consuming client).
Setting the Session Session.CLIENT_ACKNOWLEDGE will allow you to do this but then you will have to manually send an acknowledge in your consumer code. By calling acknowledge on the message msg.acknowledge() inside your onMessage() method within your listener.
This will then acknowledge the receipt of messages consumed within that session and remove them from the queue.
Pages 46 and 65 in the pdf you quoted are useful for more information on this as is the api