Exporting RMI Service in Spring framework

[JavaBean]

First write Interface and Implementation Class you don’t need to implement 'java.rmi.Remote' interface as well no need to throw RemoteException.

Code:

public interface RmiSampleService
{ 
String getValue(String s); 
} 

public class RmiSampleServiceImpl implements RmiSampleService
{ 
public static void main(String args[]) 
{ 
System.out.println("hi from RMI Spring "); 
} 
publicRmiSampleServiceImpl() 
{ 
System.out.println("constructor rmi"); 
} 
public String getValue(String s) 
{ 
return "Hi From RMI Service " + a; 
} 
}

RmiSampleService.xml

Code:

<?xml version="1.0" encoding="UTF-8"?> 
<!DOCTYPE beans PUBLIC "-//SPRING//DTD BEAN//EN" 
"http://www.springframework.org/dtd/ spring-beans.dtd"> 
<beans> 
<bean class="org.springframework.remoting.rmi.RmiServiceExporter"> 
<property name="service"> 
<value>rmiservice</value> 
</property> 
<property name="serviceName"> 
<value>service1</value> 
</property> 
<property name="serviceInterface"> 
<value>RmiSampleService</value> 
</property> 
</bean> 
<bean id="rmservice" class="RmiSampleServiceImpl"> 
</bean> 
</beans>

Now Write Client Code

Code:

RmiSampleServiceImpl.java
 
import java.io.*; 
import org.springframework.beans.factory.*; 
import org.springframework.beans.factory.xml.*; 
import org.springframework.core.io.*; 
class RmiSampleServiceClient
{ 
public static void main(String args[]) 
{ 
try 
{ 
System.out.println("Wait. for start uo."); 
Resource res = new ClassPathResource("RmiSampleService.xml"); 
BeanFactory factory = new XmlBeanFactory(res); 
System.out.println(" Bean factory created"); 
rmservice bean1 = (rmservice)factory.getBean("rmservice"); 
String s = bean1.getValue(args[0]); 
System.out.println(s); 
} 
catch(Exception e1) 
{System.out.println(""+e1);} 
} 
}

Now you need to compile and run client you will get the output.