RE: [Xmlrpc-developers] How to quit the server?

[Dominique G. <d.g...@at...>]

In fact I had the same problem with some of our code. The only work =
around
that we found was to=20
close the listening socket. That cause the hanging code ".accept()" to
throws an exception.

Why they did not implement an accept method with a timeout, that is a =
really
good question !


Dominique Gallot


-----Original Message-----
From: Greger Ohlson [mailto:qi...@ks...]
Sent: mardi 4 d=E9cembre 2001 9:25
To: marco
Cc: xml...@li...
Subject: Re: [Xmlrpc-developers] How to quit the server?




Hi Marco,

Good question. Currently, there's no way stopping an XmlRpcServer that =
has
been started with runAsService(), as far as I know. The method hangs in =
a
ServerSocket.accept() call. I'll take a closer look at it today and see
what has to be done to make it possible to stop it from another thread.
And please feel free to contribute with a solution, anyone.

Regards,
Greger


marco wrote:
>=20
> Hi List,
>=20
>    Thanks for this great Xml Rpc Implementation!
>=20
>    i have a little question, i'm running an XmlRpc Server in an =
Applet
>    with the Sun plugIn, how can i free the serverport when leaving =
the
>    site or on an reload?
>    i get this error:
>      Serverkomponente could not start: java.net.BindException: =
Address in
use: JVM_Bind
>=20
>    destroying the applet wont work, ending the thread i build either
>    not. The Serversocket is still assigned....
>=20
> --
> Best regards,
>  marco                          mailto:th...@gm...
>=20
> _______________________________________________
> Xmlrpc-developers mailing list
> Xml...@li...
> https://lists.sourceforge.net/lists/listinfo/xmlrpc-developers

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