Publish-subscribe for broadcasting in RabbitMQ - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a fanout exchange named 'logs'.

RabbitMQ

channel.exchange_declare(exchange='logs', exchange_type=[1])

Drag options to blanks, or click blank then click option'

Adirect

Bfanout

Ctopic

Dheaders

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'direct' instead of 'fanout' exchange type.

The fanout exchange broadcasts messages to all queues bound to it.

2fill in blank

medium

Complete the code to publish a message 'Hello World!' to the 'logs' exchange.

RabbitMQ

channel.basic_publish(exchange='logs', routing_key=[1], body='Hello World!')

Drag options to blanks, or click blank then click option'

A'info'

B'broadcast'

C'logs'

D''

Attempts:

3 left

💡 Hint

Common Mistakes

Using a routing key when publishing to a fanout exchange.

For a fanout exchange, the routing key is ignored, so an empty string is used.

3fill in blank

hard

Fix the error in the code to bind a queue named 'task_queue' to the 'logs' exchange.

RabbitMQ

channel.queue_bind(queue='task_queue', exchange='logs', routing_key=[1])

Drag options to blanks, or click blank then click option'

A''

B'task'

C'logs'

D'broadcast'

Attempts:

3 left

💡 Hint

Common Mistakes

Using a non-empty routing key when binding to a fanout exchange.

When binding a queue to a fanout exchange, the routing key must be an empty string.

4fill in blank

hard

Fill both blanks to declare a durable queue and bind it to the 'logs' fanout exchange.

RabbitMQ

channel.queue_declare(queue=[1], durable=[2])
channel.queue_bind(queue=[1], exchange='logs', routing_key='')

Drag options to blanks, or click blank then click option'

A'task_queue'

BTrue

CFalse

D'logs_queue'

Attempts:

3 left

💡 Hint

Common Mistakes

Using different queue names in declare and bind.

Setting durable to False when persistence is needed.

The queue name is 'task_queue' and it is declared durable to survive RabbitMQ restarts.

5fill in blank

hard

Fill all three blanks to create a callback function that prints received messages, start consuming from 'task_queue', and acknowledge messages.

RabbitMQ

def callback(ch, method, properties, body):
    print([1])
    ch.basic_ack(delivery_tag=[2])

channel.basic_consume(queue='task_queue', on_message_callback=callback, auto_ack=[3])
channel.start_consuming()

Drag options to blanks, or click blank then click option'

Abody.decode()

Bmethod.delivery_tag

CFalse

DTrue

Attempts:

3 left

💡 Hint

Common Mistakes

Forgetting to decode the message body.

Using auto_ack=True which disables manual ack.

The callback prints the decoded message body, acknowledges the message using delivery_tag, and disables auto-acknowledge to manually ack.

Practice

1. What is the main purpose of using a fanout exchange in RabbitMQ's publish-subscribe pattern?

easy

A. To broadcast messages to all queues bound to the exchange

B. To route messages to a single queue based on a key

C. To store messages until a consumer is ready

D. To filter messages based on content

Solution

1. Step 1: Understand fanout exchange behavior

   A fanout exchange sends messages to all queues bound to it without filtering or routing keys.

2. Step 2: Compare with other exchange types

   Other exchanges like direct or topic route messages selectively, but fanout broadcasts to all.

3. Final Answer:

   To broadcast messages to all queues bound to the exchange -> Option A

4. Quick Check:

   Fanout = Broadcast all [OK]

Hint: Fanout means send to all queues bound [OK]

Common Mistakes:

• Confusing fanout with direct exchange routing
• Thinking fanout filters messages
• Assuming fanout stores messages

2. Which of the following is the correct RabbitMQ command to declare a fanout exchange named logs?

easy

A. rabbitmqadmin declare exchange name=logs type=direct

B. rabbitmqadmin declare exchange name=logs type=fanout

C. rabbitmqadmin declare queue name=logs

D. rabbitmqadmin declare exchange name=logs type=topic

Solution

1. Step 1: Identify the correct exchange type for broadcasting

   The fanout type is used for broadcasting messages to all bound queues.

2. Step 2: Check the command syntax

   The command uses rabbitmqadmin declare exchange with name=logs and type=fanout to declare the exchange.

3. Final Answer:

   rabbitmqadmin declare exchange name=logs type=fanout -> Option B

4. Quick Check:

   Fanout exchange declaration = rabbitmqadmin declare exchange name=logs type=fanout [OK]

Hint: Fanout exchange uses type=fanout in declare command [OK]

Common Mistakes:

• Using direct or topic instead of fanout
• Declaring a queue instead of an exchange
• Incorrect command syntax

3. Given the following setup:

channel.exchange_declare(exchange='logs', exchange_type='fanout')
channel.queue_declare(queue='queue1')
channel.queue_bind(exchange='logs', queue='queue1')
channel.basic_publish(exchange='logs', routing_key='', body='Hello')

What will happen when the message is published?

medium

A. The message is sent to a default queue

B. The message is sent only to 'queue1' because of the routing key

C. The message is sent to all queues bound to 'logs', including 'queue1'

D. The message is discarded because routing_key is empty

Solution

1. Step 1: Understand fanout exchange behavior

   Fanout exchange ignores routing keys and sends messages to all bound queues.

2. Step 2: Analyze the code bindings

   Queue 'queue1' is bound to 'logs' exchange, so it will receive the message.

3. Final Answer:

   The message is sent to all queues bound to 'logs', including 'queue1' -> Option C

4. Quick Check:

   Fanout ignores routing key, sends to all bound queues [OK]

Hint: Fanout ignores routing_key, sends to all bound queues [OK]

Common Mistakes:

• Thinking routing_key matters for fanout
• Assuming message is discarded if routing_key is empty
• Believing message goes to default queue

4. You wrote this code to broadcast messages but only one queue receives them:

channel.exchange_declare(exchange='logs', exchange_type='fanout')
channel.queue_declare(queue='queue1')
channel.queue_declare(queue='queue2')
channel.queue_bind(exchange='logs', queue='queue1')
channel.basic_publish(exchange='logs', routing_key='', body='Test')

What is the likely problem?

medium

A. Queue 'queue2' is not bound to the exchange

B. The exchange type should be 'direct' instead of 'fanout'

C. The routing_key is missing

D. The queues are declared incorrectly

Solution

1. Step 1: Check queue bindings

   Only 'queue1' is bound to the 'logs' exchange; 'queue2' is not bound.

2. Step 2: Understand fanout behavior

   Fanout sends messages to all bound queues, so unbound queues won't get messages.

3. Final Answer:

   Queue 'queue2' is not bound to the exchange -> Option A

4. Quick Check:

   All queues must be bound to receive messages [OK]

Hint: Bind all queues to exchange to receive broadcasts [OK]

Common Mistakes:

• Assuming routing_key is needed for fanout
• Changing exchange type unnecessarily
• Not binding all queues

5. You want to broadcast logs to multiple services dynamically. Which approach best ensures each service gets its own queue and receives all messages without interfering with others?

hard

A. Services publish messages directly to each other's queues

B. All services share a single queue bound to a direct exchange

C. Each service creates a queue bound to a topic exchange with different routing keys

D. Each service creates a unique queue and binds it to a shared fanout exchange

Solution

1. Step 1: Understand publish-subscribe pattern

   Fanout exchange broadcasts messages to all bound queues, ideal for multiple receivers.

2. Step 2: Ensure isolation and dynamic addition

   Each service having its own queue bound to the fanout exchange ensures independent consumption and easy scaling.

3. Step 3: Evaluate other options

   Sharing a single queue causes message competition; topic exchange with routing keys is more selective, not broadcast; direct publishing is complex and error-prone.

4. Final Answer:

   Each service creates a unique queue and binds it to a shared fanout exchange -> Option D

5. Quick Check:

   Fanout + unique queues = broadcast to all [OK]

Hint: Use fanout exchange with unique queues per service [OK]

Common Mistakes:

• Using a single shared queue causing message loss
• Confusing topic exchange for broadcast
• Direct publishing to queues instead of exchange