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Application Identity ?

 
 
Andrew
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      4th Aug 2003
Hi

I need to uniquely identify a client application instance at runtime
and pass this value to a Server that processes message from the
client. This value is used to isolate, at the server, different client
instances. The trick is, however, that the method for doing this has
to be somehow backward compatible with a COM+ client.

I'm thinking that if I generate, at runtime, a GUID, and pass that to
the server then that should do the trick. The proviso is here however
that I do that only once per app instance and use the same interface
reference every time. That, I can pretty much guarentee though because
I have development control over the client application. So, how would
I generate a GUID at runtime in .NET (1.1) and COM+ (i.e. V6).

I'm being a little vague I know but to explain the whole concept will
take a while.

Any idea's?

Thanks in advance,
Andrew
 
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Christopher Bova
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      4th Aug 2003
Andrew,

To generate a GUID in .NET, use the static
System.Guid.NewGuid() method. In VB6, I think the
suggested idiom is to use the CoCreateGuid method in the
CoCreateGuid API.

Chris

>-----Original Message-----
>Hi
>
>I need to uniquely identify a client application instance

at runtime
>and pass this value to a Server that processes message

from the
>client. This value is used to isolate, at the server,

different client
>instances. The trick is, however, that the method for

doing this has
>to be somehow backward compatible with a COM+ client.
>
>I'm thinking that if I generate, at runtime, a GUID, and

pass that to
>the server then that should do the trick. The proviso is

here however
>that I do that only once per app instance and use the

same interface
>reference every time. That, I can pretty much guarentee

though because
>I have development control over the client application.

So, how would
>I generate a GUID at runtime in .NET (1.1) and COM+ (i.e.

V6).
>
>I'm being a little vague I know but to explain the whole

concept will
>take a while.
>
>Any idea's?
>
>Thanks in advance,
>Andrew
>.
>

 
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