Hi,
You may have used an example to write your code.
See the "getInitialContext" method defined in that sample. It may look like:
static public Context getInitialContext()
throws Exception
{
Hashtable h = new Hashtable();
h.put(Context.INITIAL_CONTEXT_FACTORY,"xxxxxxxxx.CNInitialContextFactory");
/// I am not advertising for anyone
h.put(Context.PROVIDER_URL,
"iiop://localhost:900");
/// here is your URL
return new InitialContext(h);
}
Benjamin
Seerangan Saminathan a �crit :
> Hi
> I'm a new comer to this list and just now i statred to work around EJB.
> I went thru the sun's EJB tutorial.According to tutorial,the client will do the
> follwoing tasks.
> 1.Locate the Home Interface
> 2.Create an Exterprise Bean Instance
> 3.Invoke a Business Method.
> ###########
> I did some RMI programming but in RMI client i have mentioned IP address.
> For Example obj = (Hello)Naming.lookup("//129.149.126.114/HS1");
> ################
>
> I have a basic doubt in STEP 1.
> How the client will locate HomeInterface?
> =====================================
> My Code(Part of the Client program)
> // Create A Demo object, in the server
> // Note: the name of the class corresponds to the JNDI
> // property declared in the DeploymentDescriptor
> // From DeploymentDescriptor ...
> // beanHomeName demo.DemoHome
> Context ctx = getInitialContext();
> DemoHome dhome = (DemoHome) ctx.lookup("demo.DemoHome");
> =======================================
> Let us assume my client and EJB servers are in different machine.
> If you say lookup("demo.DemoHome"), then how the client will identify the EJB
> server's IP address.
>
> Any explanation will be highly appreciated.
>
> Thank you
> Saminathan Seerangan
> 510-574-7898
>
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