0
votes

I am trying to implement JMS in my spring application. I have defined the JNDI name + queue name in applicationContext.xml as follows:

<bean id="emailQueueConnectionFactory" class="org.springframework.jndi.JndiObjectFactoryBean" lazy-init="true">
<property name="jndiName" value="java:comp/env/jms/<<Name of JNDI of connection factory>>" />
</bean>

<bean id="emailQueueDestination" class="org.springframework`enter code here`.jndi.JndiObjectFactoryBean" lazy-init="true">
<property name="jndiName" value="java:comp/env/jms/<<JNDI name of queue>>" />
</bean>

<bean id="emailQueueTemplate" class="org.springframework.jms.core.JmsTemplate" lazy-init="true">
<property name="connectionFactory" ref="emailQueueConnectionFactory" />
<property name="defaultDestination" ref="emailQueueDestination" />
</bean>

<bean id="emailSender" class="<<Package>>.EmailSender" lazy-init="true">
<property name="jmsTemplate">
<ref bean="emailQueueTemplate" />
</property>
</bean>

Now my controller makes a call to the emailSender bean using the following code:

ApplicationContext context = new ClassPathXmlApplicationContext("/applicationContext.xml");
EmailSender sender =(EmailSender)context.getBean("emailSender");

The exception I get is: Error 404: Request processing failed; nested exception is org.springframework.beans.factory.BeanDefinitionStoreException: IOException parsing XML document from class path resource [applicationContext.xml]; nested exception is java.io.FileNotFoundException: class path resource [applicationContext.xml] cannot be opened because it does not exist

I am loading the applicationContext.xml at serevr start-up still my code is not able to locate this file.

Can anyone please help.??

1

1 Answers

0
votes

make sure your applicationContext.xml file is in your class path then add the class path prefix, You can try some thing like this

   ApplicationContext context = new ClassPathXmlApplicationContext("classpath*:applicationContext.xml");