Hi,
I am following the examples in the Head First
EJB book regarding local interfaces. I have the Local Home interface, the Local interface, the Bean, and the Local Client. My problem is that I can only deploy this application if I put all four files in the same jar. Otherwise, the application will not deploy. And if the four files are in the same jar, then the question arises "How do I run the client?" What should I do? I am listing the files, and the app server that I am using is Sun's App Server 8 Q2:
import javax.ejb.*;
import java.rmi.RemoteException;
public interface AdviceLocal extends EJBLocalObject {
public
String getAdvice();
}
import javax.ejb.*;
import java.rmi.RemoteException;
public interface AdviceHomeLocal extends EJBLocalHome {
public AdviceLocal create() throws CreateException;
}
import javax.ejb.*;
public class AdviceBean implements SessionBean {
private String[] adviceStrings = {"Read a Book","Listen to some music"};
public void ejbActivate(){}
public void ejbPassivate(){}
public void ejbRemove(){}
public void setSessionContext(SessionContext ctx){}
public void ejbCreate(){}
public String getAdvice() {
System.out.println("Ready for advice?");
int random = (int) (Math.random() * adviceStrings.length);
return adviceStrings[random];
}
}
import javax.naming.*;
import javax.ejb.*;
public class AdviceLocalClient {
public static void main(String[] args) {
new AdviceLocalClient().go();
}
public void go() {
Object objref = null;
try {
Context initial = new InitialContext();
objref = initial.lookup("java:comp/env/ejb/Advisor");
} catch (NamingException ex) {
ex.printStackTrace();
}
AdviceHomeLocal home = (AdviceHomeLocal) objref;
AdviceLocal advisor = null;
try {
advisor = home.create();
} catch (CreateException ex) {
ex.printStackTrace();
}
System.out.println(advisor.getAdvice());
}
}
