Checking client status

Anon

We are implementing an authentication mechanism. We are using session_id, user name, $REMOTE_ADDR and $REMOTE_PORT based authentication to avoid multiple sessions of the same user. However, if the client side machine crashes, the server wouldn't know about this. If the same user now tries to log-on either from the same machine or from a different machine, we need to make sure if the earlier registered parameters against this user is still valid or not. So we planned to validate if the $REMOTE_ADDR is active and if it is, if the $REMOTE_PORT is also active. If yes, then we deny his second login attempt. If not, then we assume that his earlier session crashed and make way of the new one. We tried all socket functions but the connection to the client ip and port is not allowed. We get the message 'Connection Refused'. Is there any other way to check if there is any client application at client's IP address and port# ? Any help is greatly appreciated.

Thanks,
Giri

Anon

I don't think this approach makes sense. The client initiates a socket-level connection; it does not sit there listening on sockets for incoming connections. And HTTP is by definition stateless in terms of connections -- except for keep-alive, which is intended to use a single connection for quick bursts of serial requests such as graphics embedded in a page, and not for session-level connectivity.

Can't you just use a timeout mechanism to support the removal of a "lock" on the user?

Or give your client a less crash-prone operating system? :-)

Anon

Yes, you are right. I can always use the timeout constraint. But it is near to the solution and not complete. Say the timeout is set to 45 minutes. If a crash happens, the same user is not allowed to login to the server till this timeout period as the server fails to understand that the first session is indeed no more till the earlier session times out.

With windows OS, you never know when you get a blue screen with system crashing .. this has really made us think all other possible options.