Creating an interface on a server application

Thread view:

Enable EMail Alerts

Start New Thread

Thread rating:

Adrian - 15 Dec 2004 13:13 GMT

Hi!
How can I create an instance of an interface on a server machine which is
exactly like another one on a client machine. I tried different methods and
no results.
Thanks!

Reply to this Message

todd.mancini@gmail.com - 15 Dec 2004 14:28 GMT

You need to compile the interface into its own class library, and then
deploy that same DLL to both machines.

A prototypical remoting sample will often involve 3 projects -- the
server exe, the client exe, and a shared class library of interfaces.
(You'd then add a reference to the shared library to both of the client
and server projects.)

Reply to this Message

Sam Santiago - 15 Dec 2004 18:57 GMT

Check out the examples on GotDotNet:
http://samples.gotdotnet.com/QuickStart/howto/default.aspx?url=/quickstart/howto
/doc/Remoting/byvalue.aspx

Thanks,

Sam

Signature

_______________________________
Sam Santiago
ssantiago@n0spam-SoftiTechture.com
http://www.SoftiTechture.com
_______________________________

> You need to compile the interface into its own class library, and then
> deploy that same DLL to both machines.
[quoted text clipped - 3 lines]
> (You'd then add a reference to the shared library to both of the client
> and server projects.)

Reply to this Message

Adrian - 16 Dec 2004 08:35 GMT

I did exactly what you explained, the methods are invoked even on different
machine, but when I created some events, they aren't fired if the client is
on another machine. So, I think that the problem is that I did not create
the instance correctly.
On server application, I wrote:

ListDictionary prop = new ListDictionary();
prop.Add("port", 1235);
SoapClientFormatterSinkProvider clientProvider = new
SoapClientFormatterSinkProvider();
SoapServerFormatterSinkProvider serverProvider = new
SoapServerFormatterSinkProvider();
serverProvider.TypeFilterLevel =
System.Runtime.Serialization.Formatters.TypeFilterLevel.Full;
HttpChannel channel = new HttpChannel(prop, clientProvider, serverProvider);
ChannelServices.RegisterChannel(channel);
System.Runtime.Remoting.RemotingConfiguration.RegisterWellKnownServiceType(t
ypeof(UserServer), "User.soap",
System.Runtime.Remoting.WellKnownObjectMode.Singleton);
intUser =
(UsersLibrary.IUser)Activator.GetObject(typeof(UsersLibrary.IUser),
http://192.168.13.10:1235/User.soap);

intUser is an instance of my interface and I think there is another way to
create it.
Thanks!

Reply to this Message

Ken Kolda - 16 Dec 2004 16:04 GMT

Although it's not what your original question seemed to be asking, it looks
like what you're trying to do is to get a reference to your singleton on the
server so you can raise events which will then be handled by the clients.
The code as you have it below is an awkward and inefficient way of doing
this as essential you're making the server a client of itself. Any method
calls to the UserServer object will have to go through all the remoting
mechanisms even though the object is actually in the same process/AppDomain.

To get a reference to the actual singleton without using GetObject() on the
server, do this:

1) Declare and instantiate a static instance of your UserServer class.
2) Replace the code that calls RegisterWellKnownServiceType() with code that
instead calls RemotingServices.Marshal(), passing in your static instance
and your URI (User.soap).

Now, in your server code, you can simply reference the static variable to
get the same instance of the UserServer class as is being accessed by the
clients.

Hope that helps -
Ken

> I did exactly what you explained, the methods are invoked even on different
> machine, but when I created some events, they aren't fired if the client is
[quoted text clipped - 12 lines]
> HttpChannel channel = new HttpChannel(prop, clientProvider, serverProvider);
> ChannelServices.RegisterChannel(channel);

System.Runtime.Remoting.RemotingConfiguration.RegisterWellKnownServiceType(t

> ypeof(UserServer), "User.soap",
> System.Runtime.Remoting.WellKnownObjectMode.Singleton);
[quoted text clipped - 5 lines]
> create it.
> Thanks!

Reply to this Message

cnfr - 17 Dec 2004 07:59 GMT

Thanks Ken!
I told you that I was pretty sure that the way wasn't right.
Unfortunately, it did not work that way you explained me.

Anyway, I appreciate your promptitude

Reply to this Message

Adrian - 17 Dec 2004 08:30 GMT

I have to apologize.
I made a stupid mistake. Now it works!
Thanks a lot!

Reply to this Message

Creating an interface on a server application

Free Magazines