Open bujingyun8 opened 6 years ago
Sorry for the late reply.
You cannot directly send a request (or response) message to a client, since it will drop all unexpected messages silently.
There is this feature called Observing you may have interest in. It enables the server side to send more responses after the first one, or after a while.
In the server side,may I send a request after response. I try that after exchange.Respond(...)
Request request = Request.NewGet(); request.Destination = exchange.Request.Source; exchange.Request.EndPoint.SendRequest(request);
it donot work