Question

如何为下面给出的Java代码设置应用程序？

Hashtable<String, String> properties = new Hashtable<String, String>();
            properties.put(Context.INITIAL_CONTEXT_FACTORY, "com.test.factory");
            properties.put("com.domain", "DevelopmentDomain");
//            properties.put(Context.PROVIDER_URL, "tcp://test:0506");
            properties.put(Context.PROVIDER_URL, "tcp://10.00.0.00:0506");


            properties.put(Context.SECURITY_PRINCIPAL, "aaa");
            properties.put(Context.SECURITY_CREDENTIALS, "aaa");

            javax.naming.Context context = new javax.naming.InitialContext(properties);
            ConnectionFactory factory = (ConnectionFactory) context.lookup("ImpactPocQueueConnectionFactory");

            Connection connection = factory.createConnection();

            connection.start();

            Session session = connection.createSession(false, Session.AUTO_ACKNOWLEDGE);

            Destination queue = session.createQueue("test.producer");

我在这里和那里看到了许多例子，但没有一个给出完整的图片。

Answer 1

如果您想将此代码逐字转换为Spring配置，请考虑利用@Configuration方法：

@Bean
public Destination queue() throws JMSException, NamingException {
    return session().createQueue("test.producer");
}

@Bean
public Session session() throws JMSException, NamingException {
    return connection().createSession(false, Session.AUTO_ACKNOWLEDGE);
}

@Bean(initMethod = "start")
public Connection connection() throws JMSException, NamingException {
    return connectionFactory().createConnection();
}

@Bean
public ConnectionFactory connectionFactory() throws NamingException {
    return (ConnectionFactory) context().lookup("ImpactPocQueueConnectionFactory");
}

@Bean
public Context context() throws NamingException {
    return new javax.naming.InitialContext(properties());
}

@Bean
public Hashtable<String, String> properties() {
    Hashtable<String, String> properties = new Hashtable<String, String>();
    properties.put(Context.INITIAL_CONTEXT_FACTORY, "com.test.factory");
    properties.put("com.domain", "DevelopmentDomain");
    //properties.put(Context.PROVIDER_URL, "tcp://test:0506");
    properties.put(Context.PROVIDER_URL, "tcp://10.00.0.00:0506");
    properties.put(Context.SECURITY_PRINCIPAL, "aaa");
    properties.put(Context.SECURITY_CREDENTIALS, "aaa");
    return properties;
}

从技术上讲，你可以使用XML完成所有这些工作，但我发现这种方法更具可读性和可维护性。现在，您可以在范围内使用connectionFactory和queue个bean。您可以轻松地与Spring JMS support集成。如果您需要进一步的帮助，请告诉我们。

Answer 2

下面给出了Spring applicationContext xml配置已经过测试，并且可以正常使用。

<bean id="jndiTemplate" class="org.springframework.jndi.JndiTemplate">
        <property name="environment">
            <props>
                <prop key="java.naming.factory.initial">com.test.factory</prop>
                <prop key="com.sonicsw.jndi.mfcontext.domain">DevelopmentDomain</prop>
                <prop key="java.naming.provider.url">tcp://10.00.0.00:0506</prop>
                <prop key="java.naming.security.principal">aaa</prop>
                <prop key="java.naming.security.credentials">aaa</prop>
            </props>
        </property>
    </bean>
    <bean id="jmsQueueConnectionFactory" class="org.springframework.jndi.JndiObjectFactoryBean">
        <property name="jndiTemplate" ref="jndiTemplate"/>
        <property name="jndiName" value="ImpactPocQueueConnectionFactory"/>
    </bean>
    <bean id="sonic" class="org.apache.camel.component.jms.JmsComponent"> 
        <property name="connectionFactory" ref="jmsQueueConnectionFactory"/> 
    </bean>

    <route>
    <from uri="seda:something"/>
    <to uri="sonic:queue:queueName"/>
    </route>

Answer 3

尝试使用spring JMS Template或MessageListenerContainer。

如何在Spring的applicationContext中设置JMS配置？

3 个答案: