Home Forums Support Connection closed immediately after establishing a connection in the server

  • This topic has 0 replies, 1 voice, and was last updated 8 years ago by Anonymous.
Viewing 1 post (of 1 total)
  • Author
    Posts
  • #4421
    Anonymous
    Inactive

    I have a server side NetworkComms Application that serves login requests from the client’s application.
    In my logs, I have this output when I logged in my client application.

    Console OUTPUT:

    Connection Info: [TCP-E-E] xx.xxx.xx.xx:6061 -> xxx.xxx.xx.xx:8684 (e46ziHzQekaGYIb65i2-zw)
    [This part immediately closes]
    Connection closed with [TCP-E-S] xx.xxx.xx.xx:6061 -> xxx.xxx.xx.xx:8684 (e46ziHzQekaGYIb65i2-zw)
    [Another port is assigned to my log in session]

    Here is my code in C#:

    private static void LoginMessage(PacketHeader header, Connection connection, string message)
    {
    if (connection.ConnectionAlive())
    {
    try
    {
    Console.WriteLine(“\nA message was received from ” + connection.ToString() + ” which said ‘” + message + “‘.”);
    IPAddress ip = new IPAddress(connection);
    string[] info = message.Split(new char[] { ‘,’ });
    string user = info[0];
    string pass = info[1];
    Console.WriteLine(“user: ” + user);
    Console.WriteLine(“pass: ” + pass);
    Console.WriteLine(“ip: ” + ip.IP);
    Console.WriteLine(“port: ” + ip.Port);
    Console.WriteLine(“LOGIN: # of Connections: ” + NetworkComms.GetExistingConnection().Count);

    string login = accounts_model.login(user, pass);
    if (!login.Equals(” \”[]\””))
    {
    accounts_model.updateLogin(user, pass, ip.IP, int.Parse(ip.Port));
    connection.SendObject(“Receiver”, login);
    Console.WriteLine(“Connection based sending: “);

    }
    else
    connection.SendObject(“Receiver”, “null”);
    }
    catch (Exception e)
    {
    Console.WriteLine(“LoginMessage: ” + e.Message);
    }
    }
    }

    Why is it so that my Connection terminates as soon as it is established?

Viewing 1 post (of 1 total)
  • You must be logged in to reply to this topic.