CODEWITHSUNDEEP
springspring-integrationspring-jms
crm86
crm86

Reputation: 1394

JMS container factory manual startup

I have a Spring JMS application configured via annotations and I am trying to provide some information to the application BEFORE the JMS listeners start. After that, I want to start manually the listeners.

With the following configuration:

@Bean(name = "queueContainerFactory")
public JmsListenerContainerFactory<?> queue(ConnectionFactory cf) {
        SimpleJmsListenerContainerFactory factory = new SimpleJmsListenerContainerFactory();
        ActiveMQConnectionFactory amqCf = (ActiveMQConnectionFactory) cf;
        factory.setConnectionFactory(amqCf);
    ...
}

I saw the option: factory.setAutoStartup(FALSE);

With this the application context starts and the @JmsListener is not started but I do not know how to start the JMS container factory manually.

@JmsListener(containerFactory="queueContainerFactory", destination = "${destination}")
    public void jmsListener(String message) {
...
}

Upvotes: 2

Views: 1190

Answers (1)

Artem Bilan
Artem Bilan

Reputation: 121552

You have to autowire JmsListenerEndpointRegistry and obtain the particular MessageListenerContainer by its id. Where that id you can configure on the @JmsListener:

/**
 * The unique identifier of the container managing this endpoint.
 * <p>If none is specified, an auto-generated one is provided.
 * @see org.springframework.jms.config.JmsListenerEndpointRegistry#getListenerContainer(String)
 */
String id() default "";

That might look like:

@JmsListener(id ="myContainer", 
           containerFactory="queueContainerFactory", 
           destination = "${destination}")
    public void jmsListener(String message) {
...
}

...

@Autowired
JmsListenerEndpointRegistry jmsListenerEndpointRegistry;

...

this.jmsListenerEndpointRegistry.getListenerContainer("myContainer").start();

Upvotes: 1

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