I have a scenario in my RabbitMQ setup that I'm curious about how to solve. The diagram below illustrates it (exchanges and most queues removed for succinctness):
Scenario
Producer creates message A(1), it is received by the top consumer, which begins processing the message.
Producer creates message A(2), it is received by the bottom consumer (assuming both consumers are on a round-robin exchange).
The bottom consumer publishes message B(2), which is put into Message B consumer's queue
The poor slow top consumer finally finishes and emits its message B(1).
Problem
If we assume that B consumer cannot be made idempotent, how do we ensure the result of both B messages are applied in the correct order?
I had thought of using a timestamp that is applied to the initial publish of message A, and having the consumer maintain a timestamp of last change, rejecting any timestamps before that time, but that only works if each message causes the exact same kind of change and requires a lot of tracking.
Other ideas for how to approach this would be appreciated. Thanks!
I am not sure what is specific to RabbitMQ here, but the idea with timestamps sounds like a good start if you have a single producer.
The producer attaches a timestamp to the messages A, each message B take the same timestamp of its respective message A.
With your approach some messages would not be processed, eg, message B(1). If all messages should be processed by consumer B, but they should be processed in a deterministic order, then you can do a deterministic merge:
Consumer B is equipped with two queues, one queue for each consumer A. Consumer B always checks the top of both queues:
if both queues are non-empty, consumer B pops the message with the lowest timestamp.
if at least one queue is empty, the consumer B waits.
With this approach the order in which consumer B processes messages is given by the timestamps of the producer and no message is discarded. Assumptions are:
queues are FIFO
no process crashes
always the case that eventually each consumer A processes a message
consumer B can check the top of the queues in a non-blocking fashion
Related
There is a single consumer that receives all messages from a particular queue. The source of the messages is a single publisher. In my case it is a one-way communication and it is important to deliver all messages in order without losses. How can the consumer know that some messages have been dropped by RabbitMQ, for example, because the queue size limit has been reached?
I know that a message sequence number (defined by RabbitMQ) is used to confirm delivery to the consumer, so the question is: Is the sequence number a signal that messages have been dropped? For example, if the sequence number is not linear (difference between the received message and the previous message is greater than 1).
A producer sends the message with routing key (x) to the exchange once in 5 secs. Based on the routing key, it is send to 2 queues A and B. The consumer which consumes from A wants it once in 5 secs , but the consumer that consumes from B needs the message only once in one min..
One way is to consume the messages from Queue B and discard the messages that we don't need(only considering once in one min) .
Is there any other better way for this to do in rabbitmq ?
A consumer can keep state about the last time he consumed a message. When a new message arrives, can check if the the desired time has passed since the last message was consumed. If it wasn't, he can just ignore the message.
This is logic that must be handled by the consumer, not the RabbitMQ broker.
The deduplication exchange of the RMQ deduplication plugin has been designed to fit this purpose.
You can specify the time in which you don't want to see a similar message more than once and the exchange will drop any further copy of the given message. Messages are identified via the x-deduplication-header.
The amount of time a message is guaranteed to be unique can be controlled via the x-cache-ttl exchange argument or message header.
Is there is a way by which we can restrict RabbitMQ Queue to dispatch only a fixed number of messages from the Queue to the consumers?
I have 2 Queues Q1 and Q2 and 10 consumers.Every consumer can process the messages from Q1 and Q2.At any given time, only 2 consumers should process messages from Q2.All the 10 consumers can process message from Q1 simultaneously.
Is there any configuration in RabbitMQ which we can specify, so that RabbitMQ pushes only 2 messages from Q2 to any free consumer and push the next 2 only after they are acknowledged, even though other consumers are free and ready to consume.
More background on the issue:
Why only process 2 messages at a time ? :
Q2 messages are doing a web service call and the web service end point(third party) can only service 2 messages concurrently.
Cant we use concurrency ? :
If we use a ListenerContainer (Spring AMQP) the container is per consumer. We can restrict how many message one consumer can take at a time, but when we have 10 consumers, if there are messages in the Queue, each consumer will get its share.
Can we configure only 2 consumers listening to Q2 ? :
I understand we can achieve this by configuring only 2 consumers for Q2, but I am trying to avoid that. If for some reason these 2 consumers goes down, the processing of Q2 will be halted. If 10 consumers are configured, we can guarantee the processing will happen until the last consumer is down.
Looking to see if there is some config in RabbitMQ which we can make use of or any suggested solution.
Thanks in advance !
I'm pretty sure that consumer prefetch will accomplish what you want. But, Q2 can only have one consumer for this to work. There is no way to coordinate among multiple consumers - you would have to do that yourself, and could use RabbitMQ to do the coordination.
NOTE: the RabbitMQ team monitors the rabbitmq-users mailing list and only sometimes answers questions on StackOverflow.
I think you're getting wrapped up in the problem definition. What you really need is trivial, so let's break this down a bit.
Given two queues, Q1 and Q2
10 consumers
Every consumer can process the messages from Q1 and Q2.
At any given time, only 2 consumers should process messages from Q2.
All the 10 consumers can process message from Q1 simultaneously.
Comments on problem statement
First, queues are assumed to be indepenedent. An independent process P will have queue Q, thus Q1 serves process P1. This is a strict mathematical requirement - you cannot define two queues for a single process P.
Thus, the second constraint is mathematically incorrect, for the same reason that you could not write a valid function that accepts a parameter of type string and bool interchangeably. It must accept one or the other, as they are not compatible types, or it must accept a single common ancestor of the types without regard to the subtypes. This is a variant of the Liskov Substitution Principle.
Redefining the problem
There are a total of 12 consumers in the system:
Q1 has 10 consumers
Q2 has 2 consumers
[Important] Consumers are not shared between queues
Is there any configuration in RabbitMQ which we can specify, so that RabbitMQ pushes only 2 messages from Q2 to any free consumer and push the next 2 only after they are acknowledged, even though other consumers are free and ready to consume.
Based on the new definition of the problem, you have two options:
Use a Basic.Get - pull the next message from the queue as soon as the consumer finishes processing the last message.
Use consumer prefetch with limit 1. This will deliver the first and second messages for each consumer immediately, then deliver additional messages one at a time as the next message for that consumer is acknowledged. This is a bit more complicated, but might make sense if your latency margins are less than 10 milliseconds.
Note that by properly defining the problem space, we have eliminated the fundamental problem of trying to figure out how to ensure only two consumers are processing Q2 messages at any time.
try the new feature Single Active Consumer from version 3.8+.
Single active consumer allows to have only one consumer at a time consuming from a queue and to fail over to another registered consumer in case the active one is cancelled or dies.
Consuming with only one consumer is useful when messages must be consumed and processed in the same order they arrive in the queue.
Single active consumer can be enabled when declaring a queue, with the x-single-active-consumer argument set to true
https://www.rabbitmq.com/consumers.html#single-active-consumer
e.g. with the Java client:
I have one direct exchange. There is also one queue, bound to this exchange.
I have two consumers for that queue. The consumers are manually ack'ing the messages once they've done the corresponding processing.
The messages are logically ordered/sorted, and should be processed in that order. Is it possible to enforce that all messages are received and processed sequentially accross consumer A and consumer B? In other words, prevent A and B from processing messages at the same time.
Note: the consumers are not sharing the same connection and/or channel. This means I cannot use <channel>.basicQoS(1);.
Rationale of this question: both consumers are identicall. If one goes down, the other queue starts processing messages and everything keeps working without any required intervention.
One approach to handling failover in a case where you want redundant consumers but need to process messages in a specific order is to use the exclusive consumer option when setting up the bind to the queue, and to have two consumers who keep trying to bind even when they can't get the exclusive lock.
The process is something like this:
Consumer A starts first and binds to the queue as an exclusive consumer. Consumer A begins processing messages from the queue.
Consumer B starts next and attempts to bind to the queue as an exclusive consumer, but is rejected because the queue already has an exclusive consumer.
On a recurring basis, consumer B attempts to get an exclusive bind on the queue but is rejected.
Process hosting consumer A crashes.
Consumer B attempts to bind to the queue as an exclusive consumer, and succeeds this time. Consumer B starts processing messages from the queue.
Consumer A is brought back online, and attempts an exclusive bind, but is rejected now.
Consumer B continues to process messages in FIFO order.
While this approach doesn't provide load sharing, it does provide redundancy.
Even though this is already answered. May be this can help others.
RabbitMQ has a feature known as Single Active Consumer, which matches your case.
We can have N consumers attached to a Queue but only 1 (one) of them will be actively consuming messages from the Queue. Fail-over happens only when active consumer fails.
Kindly take a look at the link https://www.rabbitmq.com/consumers.html#single-active-consumer
Thank you
Usually the point of a MQ system is to distribute workload. Of course, there are some situations where processing of message N depends on result of processing the message N-1, or even the N-1 message itself.
If A and B can't process messages at the same time, then why not just have A or just B? As I see it, you are not saving anything with having 2 consumers in a way that one can work only when the other one is not...
In your case, it would be best to have one consumer but to actually do the parallelisation (not a word really) on the processing part.
Just to add that RMQ is distributing messages evenly to all consumers (in round-robin fashion) regardless on any criteria. Of course this is when prefetch is set to 1, which by default it is. More info on that here, look for "fair dispatch".
Assume several producers publish to the same exchange E (fanout). Each producer has its own channel. Queue Q is bound to exchange E. producer P1 publishes message M1 to E and receives acknowledge A1 from E. Only after acknowledge A1 second producer P2 publishes second message M2. Does RabbitMQ guarantie order of messages in Q: M1 is first, M2 is second? That is will subscribed to Q consumer always receive M1 and after that M2?
RabbitMQ guarantees order of messages in a queue: First In, First Out. The first message to go into the queue will be the first message to come out of the queue, and they will remain in order (assuming you are just consuming and acking them... if you start nacking / rejecting message, re-publishing them, etc, things change)
That is the only guarantee that it will make on the order of messages: FIFO Queues.
If you need to guarantee the order that messages are delivered to a queue, you have to build that process yourself.
FWIW, it's very difficult to build this guarantee. The only truly guaranteed way to ensure the order of messages is not to send the next one until after the first one has been processed.
Even if you wait for the publisher acknowledgement before sending the next one, it is possible for the next one to end up in the queue before the first one (though it is highly unlikely).
You may want to look into Message Sequence and Resequencer if you need to guarantee the client gets messages in a certain order.