Spring应用程序中使用JNDI的JMS实现
我正在尝试在我的 Spring 应用程序中实现 JMS。我在 applicationContext.xml 中定义了 JNDI 名称 + 队列名称，如下所示：
<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>
现在我的控制器使用以下代码调用 emailSender bean：
ApplicationContext context = new ClassPathXmlApplicationContext("/applicationContext.xml");
EmailSender sender =(EmailSender)context.getBean("emailSender");
我得到的异常是： 错误 404：请求处理失败；嵌套异常是 org.springframework.beans.factory.BeanDefinitionStoreException: IOException 从类路径资源 [applicationContext.xml] 解析 XML 文档；嵌套异常是 java.io.FileNotFoundException：无法打开类路径资源 [applicationContext.xml]，因为它不存在
我正在 serevr 启动时加载 applicationContext.xml，但我的代码仍然无法找到此文件。
有人可以帮忙吗？？
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.??
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确保您的 applicationContext.xml 文件位于您的类路径中，然后添加类路径前缀，您可以尝试这样的事情
make sure your applicationContext.xml file is in your class path then add the class path prefix, You can try some thing like this