>-----Original Message-----
>Even though this might work, I'd acutally recommend a
different approach
>like the one presented at
>
http://www.ingorammer.com/RemotingFAQ/REGISTER_HOW_MANY_C HANNELS.html
>
>You might otherwise run into problems when more than one
user calls your
>ASPX page at the same time.
>
>Cheers,
>-Ingo
>
>Independent .NET and Web Services Consultant
>Author of "Advanced .NET Remoting"
>Microsoft Regional Director - Austria
>
http://www.ingorammer.com >
>Read my "Distributed .NET Newsletter" at
http://www.ingorammer.com/NL >
>
>"gullit" <goolit@sina.com> wrote in message
>news:09e501c366e1$9cdb2c30$a301280a@phx.gbl...
>> I have resolve this problem.
>> Sample code is:
>>
>> BinaryClientFormatterSinkProvider mysinkprovider = new
>> BinaryClientFormatterSinkProvider();
>> TcpClientChannel tcc = new TcpClientChannel
>> ("channel_1",mysinkprovider);
>> ChannelServices.RegisterChannel(tcc);
>>
>> //do something
>> ......
>>
>> ChannelServices.UnregisterChannel(tcc);
>>
>> Now my web application can support at least two pages
to
>> call remoting service at one time!
>> (I do test:2 and 3 pages call remoting service at one
>> time,no exception has been thrown.)
>> You can try it!
>>
>> >-----Original Message-----
>> >I came accross to the same problem,in my web
>> >application,there are more then two pages to call
>> >remoting service.If as least two pages post request at
>> >one time,it will be happened than one will throw
>> >exception ,channel has been registed.
>> >>-----Original Message-----
>> >>Hi,
>> >>I am registering 2 HttpChannels in the same app, one
>> >without a custom
>> >>provider and one with. The first channel that has no
>> >custom provider
>> >>overrides the second one and does not let the custom
>> >sink be part of the
>> >>second channel. They are both registering different
>> >ports in the same
>> >>application and are also named differently. The only
>> way
>> >I could make this
>> >>work is by turning one of them to be a Tcp channel.
>> Does
>> >anyone have an
>> >>answer to this problem?
>> >>
>> >>thanks,
>> >>vineet
>>
>
>
>.