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Join Date: Feb 2007
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Light Poster
Hai Friends
Iam using RMI concept for my project, i want to connect client and server by using internet connection , but its working fine on LAN connection, it gives following Exception
125.17.11.229 is my internet IP
my code
plz help me for this problem...
Iam using RMI concept for my project, i want to connect client and server by using internet connection , but its working fine on LAN connection, it gives following Exception
java.lang.RuntimeException: java.rmi.ConnectException: Connection refused to hos
t: 125.17.11.229; nested exception is:
java.net.ConnectException: Connection refused: connect
at Client.init(Client.java:23)
at sun.applet.AppletPanel.run(AppletPanel.java:417)
at java.lang.Thread.run(Thread.java:619)
Caused by: java.rmi.ConnectException: Connection refused to host: 125.17.11.229; n
ested exception is:
java.net.ConnectException: Connection refused: connect
at sun.rmi.transport.tcp.TCPEndpoint.newSocket(TCPEndpoint.java:601)
at sun.rmi.transport.tcp.TCPChannel.createConnection(TCPChannel.java:198
)my code
Server:
import java.rmi.registry.LocateRegistry;
import java.rmi.registry.Registry;
import java.rmi.server.UnicastRemoteObject;
public class Server implements RemoteInterface
{
public String getMessage()
{
return "Hello World";
}
public static void main(String args[])
{
try
{
String REGISTRY_NAME = "RMI_Example";
int REGISTRY_PORT = 3032;
Registry registry = LocateRegistry.getRegistry(REGISTRY_PORT);
RemoteInterface remoteReference =
(RemoteInterface) UnicastRemoteObject.exportObject(new Server());
registry.rebind(REGISTRY_NAME, remoteReference);
}
catch (Exception e)
{
throw new RuntimeException(e);
}
}
}
Interface:
import java.rmi.Remote;
import java.rmi.RemoteException;
public interface RemoteInterface extends Remote
{
String getMessage() throws RemoteException;
}
client:import java.rmi.registry.LocateRegistry;
import java.rmi.registry.Registry;
import javax.swing.JApplet;
import javax.swing.JLabel;
import java.rmi.*;
public class ClientApplet
{
public static void main(String arg[])
{
String REGISTRY_NAME = "RMI_Example";
int REGISTRY_PORT = 3032;
try
{
Registry registry =
LocateRegistry.getRegistry("125.17.11.229",REGISTRY_PORT);
RemoteInterface remoteReference =
(RemoteInterface) registry.lookup(REGISTRY_NAME);
System.out.println(remoteReference.getMessage());
}
catch (Exception e)
{
throw new RuntimeException(e);
}
}
} Last edited by ~s.o.s~ : Apr 30th, 2008 at 4:01 pm. Reason: Added code tags, learn to use them.
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Join Date: Feb 2006
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Firewall?
Java Programmer and Sun Systems Administrator
----------------------------------------------
Debugging is twice as hard as writing the code in the first place. Therefore, if you write the code as cleverly as possible, you are, by definition, not smart enough to debug it.
--Brian Kernighan
----------------------------------------------
Debugging is twice as hard as writing the code in the first place. Therefore, if you write the code as cleverly as possible, you are, by definition, not smart enough to debug it.
--Brian Kernighan
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Join Date: Nov 2004
Location: Netherlands
Posts: 5,752
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either that or the server process is bound to the local network address and therefore trying to call the process through its external address isn't going to get a response.
As people are clearly allowed to attack me but I'm not allowed to defend myself, I no longer post to this site.
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Join Date: Feb 2007
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Light Poster
When iam using external ip(125.17.11.232) to connect server its working fine in LAN but it gives error when iam try to run in internet connection
Exception in thread "main" java.lang.RuntimeException: java.rmi.ConnectIOExcepti
on: Exception creating connection to: 194.9.30.245 nested exception is:
In error show my server local ip 194.9.30.245 ,
coding:
Server :
Server.java
import java.rmi.registry.LocateRegistry;
import java.rmi.registry.Registry;
import java.rmi.server.UnicastRemoteObject;
public class Server implements RemoteInterface
{
public String getMessage()
{
return "Hello World";
}
public static void main(String args[])
{
try
{
String REGISTRY_NAME = "RMI_Example";
int REGISTRY_PORT = 60020;
Registry registry = LocateRegistry.getRegistry(REGISTRY_PORT);
RemoteInterface remoteReference =
(RemoteInterface) UnicastRemoteObject.exportObject(new Server());
registry.rebind(REGISTRY_NAME, remoteReference);
}
catch (Exception e)
{
throw new RuntimeException(e);
}
}
}
--------------------------------------
Interface
RemoteInterface.java
import java.rmi.Remote;
import java.rmi.RemoteException;
public interface RemoteInterface extends Remote
{
String getMessage() throws RemoteException;
}
-----------------------------------------------------------
Client
import java.rmi.registry.LocateRegistry;
import java.rmi.registry.Registry;
import javax.swing.JApplet;
import javax.swing.JLabel;
import java.rmi.*;
public class ClientApplet
{
public static void main(String arg[])
{
String REGISTRY_NAME = "RMI_Example";
int REGISTRY_PORT = 60020;
try
{
Registry registry =
LocateRegistry.getRegistry("125.17.11.232",REGISTRY_PORT);
RemoteInterface remoteReference =
(RemoteInterface) registry.lookup(REGISTRY_NAME);
System.out.println(remoteReference.getMessage());
}
catch (Exception e)
{
throw new RuntimeException(e);
}
}
}
Please send me solution for this problem
Exception in thread "main" java.lang.RuntimeException: java.rmi.ConnectIOExcepti
on: Exception creating connection to: 194.9.30.245 nested exception is:
In error show my server local ip 194.9.30.245 ,
coding:
Server :
Server.java
import java.rmi.registry.LocateRegistry;
import java.rmi.registry.Registry;
import java.rmi.server.UnicastRemoteObject;
public class Server implements RemoteInterface
{
public String getMessage()
{
return "Hello World";
}
public static void main(String args[])
{
try
{
String REGISTRY_NAME = "RMI_Example";
int REGISTRY_PORT = 60020;
Registry registry = LocateRegistry.getRegistry(REGISTRY_PORT);
RemoteInterface remoteReference =
(RemoteInterface) UnicastRemoteObject.exportObject(new Server());
registry.rebind(REGISTRY_NAME, remoteReference);
}
catch (Exception e)
{
throw new RuntimeException(e);
}
}
}
--------------------------------------
Interface
RemoteInterface.java
import java.rmi.Remote;
import java.rmi.RemoteException;
public interface RemoteInterface extends Remote
{
String getMessage() throws RemoteException;
}
-----------------------------------------------------------
Client
import java.rmi.registry.LocateRegistry;
import java.rmi.registry.Registry;
import javax.swing.JApplet;
import javax.swing.JLabel;
import java.rmi.*;
public class ClientApplet
{
public static void main(String arg[])
{
String REGISTRY_NAME = "RMI_Example";
int REGISTRY_PORT = 60020;
try
{
Registry registry =
LocateRegistry.getRegistry("125.17.11.232",REGISTRY_PORT);
RemoteInterface remoteReference =
(RemoteInterface) registry.lookup(REGISTRY_NAME);
System.out.println(remoteReference.getMessage());
}
catch (Exception e)
{
throw new RuntimeException(e);
}
}
}
Please send me solution for this problem
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Join Date: Feb 2007
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Light Poster
any solution for this problem...
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Join Date: Feb 2006
Posts: 1,536
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Maybe, if you were to show what the nested exception is.
Java Programmer and Sun Systems Administrator
----------------------------------------------
Debugging is twice as hard as writing the code in the first place. Therefore, if you write the code as cleverly as possible, you are, by definition, not smart enough to debug it.
--Brian Kernighan
----------------------------------------------
Debugging is twice as hard as writing the code in the first place. Therefore, if you write the code as cleverly as possible, you are, by definition, not smart enough to debug it.
--Brian Kernighan
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Join Date: Feb 2007
Posts: 33
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Light Poster
java.rmi.ConnectIOException: Exception creating connection to: 194.9.30.242; nested exception is:
java.net.NoRouteToHostException: No route to host: connect
this is my Exception
java.net.NoRouteToHostException: No route to host: connect
this is my Exception
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Join Date: Feb 2006
Posts: 1,536
Reputation:
Rep Power: 10
Solved Threads: 141
That is, seemingly, not a publicly available ip address (i.e. can't be reached from the internet, at all), or you are attempting it to access throught the wrong interface on your computer, and so have no network path to it.
Java Programmer and Sun Systems Administrator
----------------------------------------------
Debugging is twice as hard as writing the code in the first place. Therefore, if you write the code as cleverly as possible, you are, by definition, not smart enough to debug it.
--Brian Kernighan
----------------------------------------------
Debugging is twice as hard as writing the code in the first place. Therefore, if you write the code as cleverly as possible, you are, by definition, not smart enough to debug it.
--Brian Kernighan
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