Exporting RMI Service in Spring framework

[Java Tip]

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

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

<?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

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.