RabbitMQ Message Exchange - c#

I have Created a RabbitMQ Producer and a RabbitMQ Consumer....
suppose my producer produces 10 messages. How can i get a particular message from those 10 messages.
I want to know how can i uniquely identify a message and read that or consume that message.

There are several ways to do this, but the one I use most is to use a routing key that is unique to the type of message. Consumers, then, bind to that exchange using a specific routing key, which causes messages to go only to those consumers.
If you can avoid it, you should never just dump messages into a single queue and let the consumers sort them out. The routing keys and exchanges are powerful tools made specifically for routing messages. You should leverage that.

I have an example that shows how to do a topic queue in C# which appears to be what your looking for RabbitMQ Tutorial I also have one that shows how to use the EventingBasicConsumer to avoid blocking when getting messages RabbitMQ EventingBasicConsumer

Related

How to interop with Azure Service Bus topics created by MassTransit using Rebus?

I have an existing system that publishes messages via MassTransit to some Azure Service Bus topic named CompanyName.SystemName.Messages.SomethingGotUpdated. How can I configure a new system, using Rebus, to be a subscriber to that particular topic, then route all the messages to the appropriate IHandleMessages<SomethingGotUpdated> handler?
I looked at the documentation but couldn't really figure out how to accomplish this. Any help is greatly appreciated!
You should do it by creating a MassTransit-based receiver, that receives the messages from the given topic, and then forwards the messages (by using Rebus' one-way client) to your Rebus endpoint(s).
The reason why you should do it like this, is that MassTransit and Rebus are most likely using Azure Service Bus in slightly different ways, and with very different headers on the messages, meaning that the two service buses will not be readily wire-compatible.
Some people might feel an urge to somehow map their way out of problems like this, probably by inserting a step into the incoming messages pipeline that maps MassTransit headers to Rebus headers, but I fear that you will not even have your step executed because Rebus requires a specific message ID header to be present, otherwise it will immediately move the message to the error queue.
My suggestion is to consider your MassTransit-based system thing of its own, as if it was using a completely separate network for communicating, and then consider your Rebus-based parts a system of its own too.

RabbitMQ - Topic Exchanges and Competing Consumers

I successfully setup a Topic Exchange, and I'm able to deliver messages to several consumers at once.
I'd also like to deliver messages to competing consumers and keep using topic exchanges. I read that using the same queue name enables consumers to compete for messages. I might be mistaken, however, since I cannot make it work.
Setup for several listeners to the same topic:
Declare topic exchange
For each listener, declare a new queue with an autogenerated name
Bind this queue to the above exchange with a given topic routing key
How to setup competing consumers to the same topic?
Is it even possible with Topic Exchanges?
Thanks.
Let's review a couple of points first.
First, remember that in RabbitMQ you always consume from queues. Exchanges are just your portals and you cannot directly consume from them.
Second, Topic exchanges allow for binding the queues with routing key "patterns". Therefore, the term topic is valid in the context of "Topic Exchanges".
Now this is what I understand from your question:
Multiple consumers/same routing key:
This is where you want multiple consumers to all consume the messages with the same routing key (or same routing key patterns in the case of Topic Exchanges). This is in fact doable. Just do this:
Declare your Topic Exchange
Declare a queue with some name
Bind that queue to your topic with your desired routing key pattern
Create multiple consumers and have them listen to that same queue.
What will happen is that RabbitMQ is going to load balance to your consumers in a round robin matter. This means all consumers will consume from the same queue. But remember that in this scenario it is possible that a single message is delivered more than once in theory.
What you were doing was to create multiple queues and have one consumer per queue. This means that every message coming to the exchange would be duplicated across all queues. The end result would be that a message gets processed multiple time.
I solved this by using exchange-to-exchange bindings.
The outer exchange is a topic exchange.
The inner exchange is a fanout exchange bound to a client-named queue.
The outer exchange is bound to the inner exchange with a routing key that includes a wildcard.

How to Dequeue an Enqueued Message?

I've been working on sending and receiving ActiveMQ messages between .net applications and everything work great when there is a active Consumer Connection.
However there are scenarios that I won't be able to have a active Consumer, example: web application consuming active messages.
I notice that any message not delivered to a consumer is place in a "Messages Enqueued" bucket.
Is there a way I can consume an already send and Enqueued message for a particular topic?
I am workign with Topic only, as shown on the example below
any tips from .net or even java might help me out thanks
Topics (except for durable topics) are inherently going to throw messages away if you don't have a consumer connected at the time the message is sent. That's just how topics work. My guess is that the web UI's column header is misleading- it probably should say "messages sent". You cannot retrieve them.
More on durable topics and queues here: http://activemq.apache.org/how-do-durable-queues-and-topics-work.html

Mixing of Pub/Sub with workqueues in RabbitMQ

I am evaluating using RabbitMQ as message queue/message bus and have been looking at the example tutorials on the RabbitMQ page.
I am looking for a specific scenario not covered by the tutorials and I am not sure if and how it would be possible to do via RabbitMQ.
The setup:
Let's assume I got a service, let's call it "purchase orders" and I have to other services called "logistics" and "accounting".
When an order is sent, I want to send it as a message via RabbitMQ.
There 2 "account" and 3 "logistic" services
What would be the correct way to ensure that "account" and "logistic" will process the message only once? Using pub/sub will cause the messages to be processed twice (account) or trice (logistics) if i understand it correctly.
With work queues and prefetch=1 it would assure that only one gets it, but I have 2 services and want each type of service to get one.
Is there a way to combine both and have a work queues for each of the service, without sending 2 separate events/messages to two different exchanges?
Using pub/sub will cause the messages to be processed twice (account) or trice (logistics) if i understand it correctly.
you probably have 1 queue per worker, based on your description, and you are routing the message to all worker queues. therefore, each worker gets a copy of the message, because you routed the message to all of the queues.
what you want is a single "account" queue and a single "logistic" queue. you will have multiple account services reading from the single account queue; same for the logistic service / queue.
setting prefetch=1 is important as well. this prevents you from reading too many messages in to a single worker, at once.
Is there a way to combine both and have a work queues for each of the service, without sending 2 separate events/messages to two different exchanges?
yes - don't use a fanout exchange. use a topic or direct exchange, and use multiple routing keys to route a single message to both the account and logistics queues.
What would be the correct way to ensure that "account" and "logistic" will process the message only once?
there is no way to guarantee this, 100%. at some point, even with a proper setup like I've described, you will have a network failure or a worker crash or some other problem and a message will get processed twice. you have to account for this in you design, using some form of idempotence in your message processing.
hope that helps!

Request-response pattern with worker queues in RabbitMQ

I'm using RabbitMQ for the following scenario. When a user uses a premium search feature, I send a message via RabbitMQ to one of a few server instances. They run the same routine (DB queries and billing). I want to make sure I don't process the same message more than once.
I've come across this great tutorial but the exchange type presented in it is "Topic", which does not work for me, because I process the same message more than once.
How can I implement the request-response pattern with worker queues in RabbitMQ so that each message is handled only once and there's load balancing?
Anton Gogolev's comment above is correct. You cannot guarantee a message will be processed only once, for many reasons. But, this is often a requirement of systems - to only produce the desired result once.
The way to do that is through idempotence - the idea that no matter how many times a given message is processed, it will only make the desired change once.
There are a lot of ways to do this. One simple example is to use a shared database that tracks which messages have been processed. When you receive a message, you check to see if it has been processed already. If not, you process it. If it has, you just ignore it and move on.
In your case, if you are doing request/response and want load balancing, you probably want multiple consumers on the same queue. You could have 2 or 10 or 300 instances of your request handler listening to the same queue, and you won't have too much worry about duplicate processing.
RabbitMQ will send a given message to a single consumer. It will wait for that consumer to say it is done processing, or if the consumer crashes or rejects the message, it will requeue the message for another consumer to try again.
In this way, you will generally have only 1 request handler per request. But it will always be possible for more than one to handle the same message, which is why idempotence is important.
Regarding the use of a topic exchange vs any other type of exchange - it doesn't make much difference. There will always be the possibility of more than one queue receiving the message that you are sending, because you can have multiple queues bound to the same exchange with the same binding keys.

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