I saw here that they used .QueueUnbind upon subscription close. (using .NET API)
I started to wonder what will happen if there are multiple instances. Tried to locate what does .QueueUnbind code in this repository but no luck (too many circles).
So my question is, is that only going to unbind if there is no consumer left? Or is it going to unbind it even there is an existing active consumer?
According to comment here
Bindings are orthogonal to consumers. You can delete all bindings
without cancelling any consumers. Their queues won't see any more
messages routed to them but it doesn't cancel consumers or in any way
contradict the fact that they are still around.
If a consumer has to stop but the queue is not going away, deleting
any bindings set up by that consumer [app] is a good idea but isn't a
requirement. As long as you don't leak bindings do whatever you think
make sense. Of course deleting a queue will delete all its bindings.
Related
Saw similar questions but different expected answers. My question is can I can create a consumer to focus on a single queue until it empties then switch to the other queue, until new work is sent to their main queue?
For example: 1 queue contains large amount of work to be processed in longer time frame and its own dedicated consumers (3 for instance). The 2nd queue receives much less work that requires less processing. If the consumers for the 2nd queue complete their work can I make it so they switch to the first queue until their queue receives more work?
I think for this question, it's important to keep in mind that there is a difference between a "consumer" in the canonical sense vs. a "consumer" in the RabbitMQ sense.
A RabbitMQ Consumer is a contrivance of the protocol - basically, it is a designation that the channel/connection would like to have messages pushed to it, under a designated consumer tag. In this sense, it is merely a notification to the broker to immediately send messages.
In the canonical sense, a message consumer is any piece of code that processes messages.
So, the answer to your question is "yes, go ahead and write your program to do that." You have control over the canonical consumer code. It is up to your software to determine what to do with a message that arrives from a queue.
Now, if you're wondering if RabbitMQ can re-subscribe a consumer to a different queue, the answer is "that's not how it works." In RabbitMQ, a consumer is simply a response to a request to subscribe to a queue - it is a "consumer tag" object. The ongoing nature of the subscription is tied to the channel/connection pair.
What should you do? While your code doesn't specify any particular coding language, in my opinion, you're off-track by even asking this question. Subscribe to both queues. If there is nothing for the worker to do, I think the computer would be perfectly happy with that. If you're worried about a particularly busy queue issuing too much work, you can use a number of techniques to throttle messages coming into that consumer. One popular technique is prefetch.
In my app(multiple instances), we occasionally see the case where connection is lost between my app and rabbitmq due to network issues(my app and rabbitmq are both alive), then after connection is recovered(re-established) we will receive messages that are unacked.
This creates an issue for us, because my app wasn't dead, and it is still processing the same message it received before, but now the message is redeivered, and it causes the app to process the message again (which can be fatal to us).
Since the app has multiple instances, it is not easy for an instance to check if another instance is processing the same message at the same time. We can't simply filter out redelivered message, because we need this feature to handle instance/app crashes/re-deployments.
It doesn't seem that there is an api to tell rabbitmq when to not redeliver unacked messages.
So what is the recommended practice to handle this situation ?
Thanks,
The general solution for such scenario is to make the consumers handle the messages in an idempotent manner . Generally what I do is from the producer side ( in case there is no unique identifier in the message body ) I add an attribute idempotencyId to the message body which is a guid and on the consumer side for each message this id is validated against the stored value in database , any duplicates are rejected.
This approach also works for messages which might be shoveled from another cluster or if in a same cluster multiple instances of consumers are listening then too this approach guarantee one time processing.
Would suggest to go over the RabbitMQ Reliability Guide here
Yeah, exactly-once delivery is not something RabbitMQ is good at. In fact, I'd say you should probably not be using it for these kinds of problems. Honestly, the only way to truly fix this is to use distributed transactions or locking.
Anyway, you could turn the problem on its head by ack'ing the message as soon as the consumer gets it, before it starts working on it. That would avoid the RabbitMQ-related duplication issue at least. This is at-most-once delivery.
Of course, it means that if the consumer crashes, the message is lost forever. So you need to persist the message right before you ack it so you can recover it later and also the consumer should remove it once it's complete.
Considering that crashes are rare, you can then have a single dedicated process that just works on those persisted messages. Or for that matter, handle them manually.
Just be aware that you are pushing the duplication problem in front of you, because the consumer might fail to remove the persisted message after it's done working with it anyway, but at least you have the option to implement it however you want.
Storage in this case could be anything from files, a RDBMS or something like ZooKeeper or Redis to lock/unlock in-flight messages.
I have an queue and messages in it. Also i have two consumer in separate processes. I take message by one, and decide that this message is not mine, and reject it with requeue flag. In documentation I found the next phrase "The server MUST NOT deliver the message to the same client within the context of the current channel". Is that mean that the rejected message should be deliver to another consumer or not?
So, there are a couple of things going on here that I'd like to touch on.
First, your question as to the behavior of RabbitMQ. The rule referenced above comes from the AMQP-0-9-1 specification. As with most implementation of open specs, RabbitMQ is not fully-conforming. This page describes in precise detail exactly which portions of the specification are implemented, and where any deviations occur.
On that page, it stipulates that "No attempt is made to prevent redelivery to the same client." RabbitMQ lists this as a planned addition in a future release, but it has been planned for quite a few years now.
Should Consumers Be Picky?
The more important question is the one you haven't directly asked, but that is "should my consumer be picky about which messages from the queue it processes?
The answer to this is a definitive "no." One of the key design assumptions about message queues is that any consumer subscribed to the queue should be able to process any message in the queue. Thus, it should be considered proper design that all consumers attached to the queue are running identical code (same code base, same version). If not, you're going to have some serious problems with your application sooner or later.
Reject should only be used to tell the broker that there is a problem with a particular message. If there is a problem with a particular consumer (e.g. loses connection to a database), it should not reject the message, but instead should close the connection, triggering redelivery to another, working consumer. By design, messages that need to be processed by a specialized or different consumer should be deposited in a different queue.
I am trying to set up broadcast messaging to all nodes in the system. When a new node joins the system, it publishes a message to everyone else to announce its entry. The way I have designed is that, a exchange exists to which all nodes will bind its own queue. Whenever a new node joins the system, it will bind its queue as well to the exchange and publish a message to the exchange. All nodes will receive this msg(including itself) and all other nodes(except this message) will send a "ack" message so that the new node will get to know the available nodes in the system. But somehow I couldn't able to get this working. My broadcast messages doesn't propagate to every node in the system. A simple one node publishing and rest consuming is working. But same node publishing and consuming is somehow screwed up somewhere.
Is there any other efficient way of doing this apart from the logic mentioned above? Or is there any restriction from rabbitmq perspective to achieve the above or is my code buggy and do I have to take a closer look at it.
The way you described it, your solution should work. However, without more detailed code examples (of the consume/publish logic in the "announcer" and the consume/acknowledge-publish logic in the other peers) it's difficult to debug.
A couple common problems could be tripping you up, though:
If you're considering "did I get responses back from all the other nodes" as the authority for "did the other nodes get my announce message?", you might need to acknowledge (basic.ack in AMQP) the messages your announcer is receiving as it gets them. Otherwise, it's possible you're not seeing subsequent messages due to consumer prefetch, though in most client libraries you'd have to be explicitly turning that on somewhere first.
Make sure your other peers (the ones receiving the "announce" and sending a message back) are acknowledging the message as well, or are consuming in "no-ack" mode. Otherwise, if they get blocked (via flow, rate-limiting, or prefetch), they will probably receive announces for awhile and then stop.
Make sure you're using a "fanout" type exchange. It sounds like you want unconditional-fanout behavior, so you don't need to muck about with topic routing. If you're using a topic or direct exchange, you may have a bug in your routing logic, in which case switching to fanout will work. I suspect you're already doing this though.
This is likely not the issue, but: you mention that your peers (not the announcer) are "acknowledging" the announce. Make sure that they acknowledge the announce by publishing a new message back to the announcer's queue directly (with no exchange, just a routing key), not by sending a basic.ack to RabbitMQ (that doesn't notify the sender of anything), and not by publishing an announce-received to the fanout exchange.
As an aside, I don't know why you're doing declare-queue/bind/publish as opposed to publish/declare-queue/bind; is there a good reason you need an announcing node to receive its own announce message? If you're after a "self-test" behavior, I think it's probably better to just implement a periodic "can things announce successfully?" health-check somewhere instead, though that's entirely subjective.
Have you tried the RPC style message, with a callback queue that you identify in the broadcast message's propeties? Like at the rabbitmq tutorial.
Pretty new to RabbitMQ and we're still in the investigation stage to see if it's a good fit for our use cases--
We've readily come to the conclusion that our desired topology would have us deploying a few topic based exchanges, and then filtering from there to specific queues. For example, let's say we have a user and an upload exchange, where the user queue might receive messages where the topic is "new-registration" or "friend-request" and the upload exchange might receive messages like "video-upload" or "picture-upload".
Creating the queues, getting them routed to the appropriate queue, and then building listeners to handle the messages for the various queues has been quite straight forward.
What's unclear to me however is if it's possible to do a fanout on a topic exchange?
I.e. I have named queues that are bound to my topic exchange, but I'd like to be able to just throw tons of instances of my listeners at those queues to prevent single points of failure. But to the best of my knowledge, RabbitMQ treats these listeners in a straight forward round robin fashion--e.g. every Nth message always go to the same Nth listener rather than dispatching messages to the first available consumer. This is generally acceptable to us but given the load we anticipate, we'd like to avoid the possibility of hot spots developing amongst our consumer farm.
So, is there some way, either in the queue or exchange configuration or in the consumer code, where we can point our listeners to a topic queue but have the listeners treated in a fanout fashion?
Yes, by having the listeners bind using different queue names, they will be treated in a fanout fashion.
Fanout is 1:N though, i.e. each task can be delivered to multiple listeners like pub-sub. Note that this isn't restricted to a fanout exchange, but also applies if you bind multiple queues to a direct or topic exchange with the same binding key. (Installing the management plugin and looking at the exchanges there may be useful to visualize the bindings in effect.)
Your current setup is a task queue. Each task/message is delivered to exactly one worker/listener. Throw more listeners at the same queue name, and they will process the tasks round-robin as you say. With "fanout" (separate queues for a topic) you will process a task multiple times.
Depending on your platform there may be existing work queue solutions that meet your requirements, such as Resque or DelayedJob for Ruby, Celery for Python or perhaps Octobot or Akka for the JVM.
I don't know for a fact, but I strongly suspect that RabbitMQ will skip consumers with unacknowledged messages, so it should never bottleneck on a single stuck consumer. The comments on their FAQ seem to suggest that RabbitMQ will make an effort to keep things chugging along even in the presence of troublesome consumers.
This is a late answer, but in case others come across this question...
It sounds like what you want is fair dispatch rather than a fan out model (which would publish a given message to every queue).
Fair dispatch will give a message to the next available worker rather than using a simple round-robin approach. This should avoid the "hotspots" you are concerned about, without delivering the same message to multiple consumers.
If this is what you are looking for, then see the "Fair Dispatch" section on this page in the Rabbit docs. A prefetch count of 1 is the key here.