Retrieving contents inside JMS Queue for within specific time inteval - java

I need to create an application wherein I have to retrieve all the elements inside the JMS queue within a given time limit.
For instance, the given the limit is 10 seconds. So every 10 seconds, the application should create a new Thread wherein the Thread is responsible for 1) connecting to the JMS queue and 2) retrieving all the messages during the time of connection.
So in 10 seconds, lets say that there were 15 TextMessages in the queue. I only want the current executing thread to retrieve those 15 TextMessages and nothing else. I'm afraid that the thread would pick up additional messages.
Is there a facility to limit how much messages a consumer can take? Maybe something feature which would let me see how much the queue contains?

One method I can think of is that you create a receiver from a session that uses CLIENT_ACKNOWLEDGE acknowledgement mode. Now start the receiver and receive the messages. Yes you will receive some additional messages. Now as you receive a message get it JMSTimestamp and see whether it belongs to the time duration your thread is interested in. If the message is as per your time requirement acknowledge it. If not do not acknowledge it in which case it will persist on the server and may be picked up by other threads looking for messages with different time stamps.
Another efficient way would be using message selector. Since JMSTimestamp is a message header and can be used in a selector you can take advantage of it. Create receiver with a selector on JMSTimestamp with your time range requirement. Only messages satisfying the selector will be received.

Related

Executing receiveMessageRequest to the same SQS FIFO queue

I have two lambda instance running at the same time and these two instances will do a short poll to the same FIFO queue with only a few seconds apart.
The first instance will receive the first 10 messages and the second instance will receive 0 message even though there are a total of 15 messages in the queue.
Why couldn't the second instance get the remaining 5 messages from the queue? Is this the expected behaviour and how can I overcome it?
Your 15 messages (most likely) all belong to the same Message Group ID. Therefore, the remaining 5 will not become available to your consumers until the first 10 are successfully processed and deleted. For FIFO Queues, this is the expected behaviour to preserve the order of messages (cheers #Michael-sqlbot for pointing in the right direction with this answer as per comments below).
Use long polling for Standard Queues. Short polling doesn't check every SQS server, therefore, it has the potential to not get all results. Long polling does check all SQS servers and will therefore get all results.

How to reconsume a rejected message later, RabbitMQ

Sometimes due to some external problems, I need to requeue a message by basic.reject with requeue = true.
But I don't need to consume it immediately because it will possibly fail again in a short time. If I continuously requeue it, this may result in infinite loop and requeue.
So I need to consume it later, say one minute later,
And I need to know how many times the messages has been requeue so that I can stop requeue it but only reject it to declare it fails to consume.
PS: I am using Java client.
There are multiple solutions to point 1.
First one is the one chosen by Celery (a Python producer/consumer library that can use RabbitMQ as broker). Inside your message, add a timestamp at which the task should be executed. When your consumer gets the message, do not ack it and check its timestamp. As soon as the timestamp is reached, the worker can execute the task. (Note that the worker can continue working on other tasks instead of waiting)
This technique has some drawbacks. You have to increase the QoS per channel to an arbitrary value. And if your worker is already working on a long running task, the delayed task wont be executed until the first task has finished.
A second technique is RabbitMQ-only and is much more elegant. It takes advantage of dead-letter exchanges and Messages TTL. You create a new queue which isn't consumed by anybody. This queue has a dead-letter exchange that will forward the messages to the consumer queue. When you want to defer a message, ack it (or reject it without requeue) from the consumer queue and copy the message into the dead-lettered queue with a TTL equal to the delay you want (say one minute later). At (roughly) the end of TTL, the defered message will magically land in the consumer queue again, ready to be consumed. RabbitMQ team has also made the Delayed Message Plugin (this plugin is marked as experimental yet fairly stable and potential suitable for production use as long as the user is aware of its limitations and has serious limitations in term of scalability and reliability in case of failover, so you might decide whether you really want to use it in production, or if you prefer to stick to the manual way, limited to one TTL per queue).
Point 2. just requires putting a counter in your message and handling this inside your app. You can choose to put this counter in a header or directly in the body.

How to send only 1 message at once after split() on inOut exchange

I'm using Camel for a while and I'm a huge admirer of its simplicity.
The use case
Given this simple route:
from("mock:some-route")
// split 1
.split().method("splitterBean", "split")
// now we have an ArrayList of n messages (let's say 10)
.to(ExchangePattern.InOut, "jms:some-thing");
If we assume that we have 10 messages after the split(), this route will immediately send 10 messages to the "to" endpoint. So jms:some-thing will receive all 10 messages at once.
The problem
--> Please note that the "out" endpoint is inOut, so we have timeouts in place when the receiver must acknowledge the message.
The application on the receiving end of jms:some-thing has to do quite some work for each message. As all 10 messages were written at the same time, the same timeout applies for all of them.
So we increased that timeout.
But one day, we will have 1000 messages and the timeout will again be to low.
What i want to archieve
I want to implement a pattern where I'll send only 1 message at once after the split, then sending the next after 1 message is acknowledged by the receiving system.
So instead of sending the 10 messages at once, I want
Send 1 message
Wait for the acknowledgment of that message
Send the next
Wait again
And so on..
How to implement such behavior?
I looked at the documentation, but none of the EIP components seem to fulfill that need?
Thanks for any input
You can have an intermediate seda queue with only one thread.
from("mock:some-route")
.split().method("splitterBean", "split")
.to("seda:your-seda-queue?waitForTaskToComplete=Always&timeout=0");
from("seda:your-seda-queue?waitForTaskToComplete=Always&timeout=0")
.to(ExchangePattern.InOut, "jms:some-thing");
By default, a seda queue will have a single consuming thread, and will block the calling thread until a consumer becomes available. More on seda details here
Saying that, your sending to a jms topic, which is really what you should be using to queue up your requests instead of a seda queue. You should look into implementing this logic asynchronously, and waiting on a reply topic rather than using a timeout.

Amazon SQS - FIFO Queue message request, inconsistent receives

I have a FIFO SQS queue, with visibility time of 30 seconds.
The requirement is to read messages as Quickly as possible and clear the queue.
I have code in JAVA in a fashion shown below ( this is just a representation of idea only, not complete code ):
//keep getting messages from FIFO and process them ASAP
while(true)
{
List<Message> messages =
sqsclient.receiveMessage(receiveMessageRequest).getMessages();
//my logic/code here to process these messages and delete them ASAP
}
In the while loop as soon as the messages are received, they are processed and removed from the queue.
But, many times the receiveMessageRequest does not give me messages (returns zero messages).
Also, the messages limitation is only 10 at a time during receive from SQS, which is already an issue, but due to these zero receives, the queues are piling up.
I have no clue why this is happening. The documentation exactly is not clear on this part (or Am I missing in terms of the configuration of the queue?)
Please help!
Note:
1. My FIFO Queue always has messages in this scenario, so there is no case of Queue having zero messages and receive request returning zero
2. The processing and delete times are also Less than the visibility timeout.
Thanks.
Update:
I have started running multiple consumers for processing the FIFO queue. Clearly, one consumer is not coping up with the inflow of messages. I shall update in few days how multiple consumers are performing. Thanks
You have to first make sure that all messages you received are deleted within VisibilityTimeout. If you are using DeleteMessageBatch for deletion make sure that all 10 messages are deleted.
Also, how did you queue messages when you enqueue them?
Order of messages are guaranteed only in a single message group.
This also means that if you set the same group id to all messages, you are limited to a single consumer so that order of messages are preserved for sure. Even if use multiple consumers, all messages that belong to a same group becomes invisible to other consumers until visibility timeout expires.

hornetq delayed redelivery for message group

I want to somehow delay messages for the whole message group.
The thing is that all messages belonging to each message group must be processed in the same order they were posted, sequentially. If one of the messages cannot be consumed - we want to delay it and also delay the remaining ones in the same message group. I do not want to block the consumer - it should be free to process messages from other groups.
How to do that?
I can't say JMS has anything nice built in support for this stuff. Everything is easier with single "stand alone" messages, but there is one thing you could try.
Do a delayed delivery for those messages (in that group).
// Send to same queue once again, but delay 60 sec
if( isGroupMarkedForRedelivery(message.getStringProperty("JMSXGroupID"))){
message.setLongProperty("_HQ_SCHED_DELIVERY", System.currentTimeMillis() + 60000);
producer.send(message); // producer sends to process queue (again).
}
Note that if you need them in the same order, then you should probably not use concurrency in sending and/or receiving. You could of course add more logic to adapt to your situation.
You probably need to make sure isGroupMarkedForRedelivery returns false for a specific group after less amount of time than the "delay".

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