ActiveMQ Override scheduled message - overriding

I am trying to implement delayed queue with overriding of messages using Active MQ.
Each message is scheduled to be delivered with delay of x (say 60 seconds)
In between if same message is received again it should override previous message.
So even if I receive 10 messages say in x seconds. Only one message should be processed.
Is there clean way to accomplish this?

The question has two parts that need to be addressed separately:
Can a message be delayed in ActiveMQ?
Yes - see Delay and Schedule Message Delivery. You need to set <broker ... schedulerSupport="true"> in your ActiveMQ config, as well as setting the AMQ_SCHEDULED_DELAY property of the JMS message saying how long you want the message to be delayed (10000 in your case).
Is there any way to prevent the same message being consumed more than once?
Yes, but that's an application concern rather than an ActiveMQ one. It's often referred to as de-duplication or idempotent consumption. The simplest way if you only have one consumer is to keep track of messages received in a map, and check that map whether you receive a message. It it has been seen, discard.
For more complex use cases where you have multiple consumers on different machines, or you want that state to survive application restart, you will need to keep a table of messages seen in a database, and query it each time.
Please vote this answer up if it helps, as it encourages people to help you out.

Also according to method from ActiveMQ BrokerService class you should configure persistence to have ability to use scheduler functionality.
public boolean isSchedulerSupport() {
return this.schedulerSupport && (isPersistent() || jobSchedulerStore != null);
}

you can configure activemq broker to enable "schedulerSupport" with the following entry in your activemq.xml file located in conf directory of your activemq home directory.
<broker xmlns="http://activemq.apache.org/schema/core" brokerName="localhost" dataDirectory="${activemq.data}" schedulerSupport="true">

You can Override the BrokerService in your configuration
#Configuration
#EnableJms
public class JMSConfiguration {
#Bean
public BrokerService brokerService() throws Exception {
BrokerService brokerService = new BrokerService();
brokerService.setSchedulerSupport(true);
return brokerService;
}
}

Related

How to deduplicate events when using RabbitMQ Publish/Subscribe Microservice Event Bus

I have been reading This Book on page 58 to understand how to do asynchronous event integration between microservices.
Using RabbitMQ and publish/subscribe patterns facilitates pushing events out to subscribers. However, given microservice architectures and docker usage I expect to have more than once instance of a microservice 'type' running. From what I understand all instances will subscribe to the event and therefore would all receive it.
The book doesn't clearly explain how to ensure only one of the instances handle the request.
I have looked into the duplication section, but that describes a pattern that explains how to deduplicate within a service instance but not necessarily against them...
Each microservice instance would subscribe using something similar to:
public void Subscribe<T, TH>()
where T : IntegrationEvent
where TH : IIntegrationEventHandler<T>
{
var eventName = _subsManager.GetEventKey<T>();
var containsKey = _subsManager.HasSubscriptionsForEvent(eventName);
if (!containsKey)
{
if (!_persistentConnection.IsConnected)
{
_persistentConnection.TryConnect();
}
using (var channel = _persistentConnection.CreateModel())
{
channel.QueueBind(queue: _queueName,
exchange: BROKER_NAME,
routingKey: eventName);
}
}
_subsManager.AddSubscription<T, TH>();
}
I need to understand how a multiple microservice instances of the same 'type' of microservice can deduplicate without loosing the message if the service goes down while processing.
From what I understand all instances will subscribe to the event and
therefore would all receive it.
Only one instance of subscriber will process the message/event. When you have multiple instances of a service running and subscribed to same subscription the first one to pick the message will set the message invisible from the subscription (called visibility timeout). If the service instance is able to process the message in given time it will tell the queue to delete the message and if it's not able to process the message in time , the message will re-appear in queue for any instance to pick it up again.
All standard service bus (rabbitMQ, SQS, Azure Serivce bus etc) provide this feature out of box.
By the way i have read this book and used the above code from eShotContainers and it works the way i described.
You should look into following pattern as well
Competing Consumers pattern
Hope that helps!

gRPC Java Client - hasNext during onNext?

I have a server-side streaming gRPC service that may have messages coming in very rapidly. A nice to have client feature would be to know there are more updates already queued by the time this onNext execution is ready to display in the UI, as I would simply display the next one instead.
StreamObserver< Info > streamObserver = new StreamObserver< info >( )
{
#Override
public void onNext( Info info )
{
doStuffForALittleWhile();
if( !someHasNextFunction() )
render();
}
}
Is there some has next function or method of detection I'm unaware of?
There's no API to determine if additional messages have been received, but not yet delivered to the application.
The client-side stub API (e.g., StreamObserver) is implemented using the more advanced ClientCall/ClientCall.Listener API. It does not provide any received-but-not-delivered hint.
Internally, gRPC processes messages lazily. gRPC waits until the application is ready for more messages (typically by returning from StreamObserver.onNext()) to try to decode another message. If it decodes another message then it will immediately begin delivering that message.
One way would be to have a small, buffer with messages from onNext. That would let you should the current message, and then check to see if another has arrived in the mean time.

Does Rebus support batch sending

Is it possible to batch Rebus messages (using Azure Servicebus) ?
The reason is that we are going to send a lot of message to save log events and want to batch up.
While old versions of Rebus did have a batch API for wrapping up multiple logical messages inside one single transport message, that functionality turned out to bring very little advantage at the expense of increased complexity in many places.
If you want to send batches of messages, I suggest you simply code your own message batch message, something like
public class BatchOfLogEvents
{
public BatchOfLogEvents(IEnumerable<LogEvent> logEvents)
{
LogEvents = logEvents.ToArray();
}
public IReadOnlyCollection<LogEvent> LogEvents { get; }
}
and then you send that and create a handler for it in the other end.
Update regarding Azure Service Bus: Please remember that Azure Service Bus has a 256 kB maximum message size (or 1MB if you're on Premium).
Also: If you have not done so already, you can probably benefit from enabling GZip compression of messages by going
.Options(o => o.EnableCompression())
in your Rebus configurations.

JMS - Cannot retrieve message from queue. Happens intermittently

We have a Java class that listens to a database (Oracle) queue table and process it if there are records placed in that queue. It worked normally in UAT and development environments. Upon deployment in production, there are times when it cannot read a record from the queue. When a record is inserted, it cannot detect it and the records remain in the queue. This seldom happens but it happens. If I would give statistic, out of 30 records queued in a day, about 8 don't make it. We would need to restart the whole app for it to be able to read the records.
Here is a code snippet of my class..
public class SomeListener implements MessageListener{
public void onMessage(Message msg){
InputStream input = null;
try {
TextMessage txtMsg = (TextMessage) msg;
String text = txtMsg.getText();
input = new ByteArrayInputStream(text.getBytes());
} catch (Exception e1) {
// TODO Auto-generated catch block
logger.error("Parsing from the queue.... failed",e1);
e1.printStackTrace();
}
//process text message
}
}
Weird thing we cant find any traces of exceptions from the logs.
Can anyone help? by the way we set the receiveTimeout to 10 secs
We would need to restart the whole app for it to be able to read the records.
The most common reason for this is the listener thread is "stuck" in user code (//process text message). You can take a thread dump with jstack or jvisualvm or similar to see what the thread is doing.
Another possibility (with low volume apps like this) is the network (most likely a router someplace in the network) silently closes an idle socket because it has not been used for some time. If the container (actually the broker's JMS client library) doesn't know the socket is dead, it will never receive any more messages.
The solution to the first is to fix the code; the solution to the second is to enable some kind of heartbeat or keepalives on the connection so that the network/router does not close the socket when it has no "real" traffic on it.
You would need to consult your broker's documentation about configuring heartbeats/keepalives.

How to subscribe for RabbitMQ notification messages?

I am developing a Qt5 server application and I am using the QAMQP library.
What I want to do is the following:
Another server should send a message whenever something about a user
should change
My server, which is distributed among multiple machines and has multiple processes per machine needs to be notified about these updates
The thing is, I am not sure about the architecture that I should build. I just know that whenever something about some user changes, the server needs to send a message to the RabbitMQ broker and all my processes that are interested in updates for that particular user should get the message. But should I create one queue per process, and bind it with a separate exchange for each user? Or maybe create in each process a separate queue for each user and bind that somehow to some exchange. Fanout exchanges come to mind, and one queue per process, I am just not sure about the queue-exchange relations even though I've spent quiet some time trying to figure it out.
Update, in order to clarify things and write about the progress
I have a distributed application that needs to be notified for product changes. Those changes happen often and are tracked by another platform. I want to get those updates in my application.
In order to achieve that, each one of my application instances creates it's own queue. Then, whenever an instance is interested in updates for a particular product it creates an exchange for that product and binds it to the queue, like this:
Exchange type : 'direct'
Exchange name : 'product_update'
Routing key : 'PRODUCT_CODE'
Where PRODUCT_CODE is a string that represents the code of the product. In the platform that track the changes, I just publish messages with the corresponding exchanges.
The problem comes when i need to unsubscribe for a product update. I am using the QAMQP library, and in the destructor of the QAMQP::Exchange there's an unconditional remove() call.
When that function is called I am getting error in the RabbitMQ log, which looks like this:
=ERROR REPORT==== 28-Jan-2014::08:41:35 ===
connection <0.937.0>, channel 7 - soft error:
{amqp_error,precondition_failed,
"exchange 'product_update' in vhost 'test-app' in use",
'exchange.delete'}
I am not sure how to properly unsubscribe. I know from the RabbitMQ web interface that I have only one exchange ('product_update') which has bindings to multiple queues with difference routing keys.
I can see that the call to remove() in QAMQP tries to delete the exchange, but since it's used by my other processes, it's still in use and cannot be removed, which I beleive is ok.
But what should I do to delete the exchange object that I created? Should I first unbind it from the queue? I believe that i should be able to delete the object without calling remove(), but I may be mistaken or I may doing it wrong.
Also, if there's a better pattern for what I am trying to accomplish, please advice.
Here's some sample code, per request.
ProductUpdater::ProductUpdater(QObject* parent) : QObject(parent)
{
mClient = new QAMQP::Client(this);
mClient->setAutoReconnect(true);
mClient->open(mConnStr);
connect(mClient, SIGNAL(connected()), this, SLOT(amqp_connected()));
}
void ProductUpdater::amqp_connected()
{
mQueue = mClient->createQueue();
connect(mQueue, SIGNAL(declared()), this, SLOT(amqp_queue_declared()));
connect(mQueue, SIGNAL(messageReceived(QAMQP::Queue*)),
this, SLOT(message_received(QAMQP::Queue*)));
mQueue->setNoAck(false);
mQueue->declare(QString(), QAMQP::Queue::QueueOptions(QAMQP::Queue::AutoDelete));
}
void ProductUpdater::amqp_queue_declared()
{
mQueue->consume();
}
void ProductUpdater::amqp_exchange_declared()
{
QAMQP::Exchange* exchange = qobject_cast<QAMQP::Exchange*>(sender());
if (mKeys.contains(exchange))
mQueue->bind(exchange, mKeys.value(exchange));
}
void ProductUpdater::message_received(QAMQP::Queue* queue)
{
while (queue->hasMessage())
{
const QAMQP::MessagePtr message = queue->getMessage();
processMessage(message);
if (!queue->noAck())
queue->ack(message);
}
}
bool ProductUpdater::subscribe(const QString& productId)
{
if (!mClient)
return false;
foreach (const QString& id, mSubscriptions) {
if (id == productId)
return true; // already subscribed
}
QAMQP::Exchange* exchange = mClient->createExchange("product_update");
mSubscriptions.insert(productId, exchange);
connect(exchange, SIGNAL(declared()), this, SLOT(amqp_exchange_declared()));
exchange->declare(QStringLiteral("direct"));
return true;
}
void ProductUpdater::unsubscribe(const QString& productId)
{
if (!mSubscriptions.contains(productId))
return;
QAMQP::Exchange* exchange = mSubscriptions.take(productId);
if (exchange) {
// This may even be unnecessary...?
mQueue->unbind(exchange, productId);
// This will produce an error in the RabbitMQ log
// But if exchange isn't destroyed, we have a memory leak
// if we do exchange->deleteLater(); it'll also produce an error...
// exchange->remove();
}
}
Amy,
I think your doubt is related to the message distribution style (or patterns) and the exchange types available for RabbitMQ. So, I'll try to cover them all with a short explanation and you can decide which will fit best for your scenario (RabbitMQ tutorials explained in another way).
Work Queue
Using the default exchange and a binding key you can post messages directly yo a queue. Once a message arrives for a queue, the consumers "compete" to grab the message, it means a message is not delivered to more than one consumer. If there are multiple consumers listening to a single queue, the messages will be delivered in a round-robin fashion.
Use this approach when you have work to do and you want to scale across multiple servers/processes easily.
Publish/Subscribe
In this model, one single sent message may reach many consumers listening on their queues. For this scenario, where you must unselectively dispatch messages to all consumers, you can use a fanout exchange. These exchanges are "dumb" and acts just like their names imply: like a fan. One thing enters and is replicated without any intelligence to all queues that are bound to the exchange. You could as well use direct exchanges, but only if you need to do any filtering or routing on the messages.
Use this scenario when you have something like an event and you may need multiple servers, processes and consumers to handle that event, each one doing a task of different nature to handle the event. If you do not need any filter/routing, use fanout exchange for this scenario.
Routing / Topic
A particular case of the Publish/Subscribe model, where you can have queues "listen" on the exchange using filters, that may have pattern matching (topics) or not (just route).
If you need pattern matching, use topic exchange type. If you don't, use direct.
When a queue "listens" to an exchange, a binding is used. In this binding, you may specify a binding key.
To deliver the message to the correct queues, the exchange examines the message's routing key. If it matches the binding key, the message is forwarded to that queue. The match strategy depends on wether you are using topic or direct exchange, as said before.
TL;DR:
For your scenario, if each process do something different with the User change event, use a single exchange with fanout type. Each class of handler declares the same queue name bound to that exchange. This relates to the Publish/Subscribe model above. You can distribute work to among consumers of the same class listening on the same queue name, even if they don't reside on the same process.
However, if all the consumers that are interested in the event perform the same task when handling, use the work queue model.
Hope this helps,

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