如果没有响应,请在30秒后关闭Socket Listener应用程序

时间:2011-02-28 13:57:59

标签: c# .net

(C#)这里我创建了一个等待客户端命中的套接字监听器应用程序。现在我的要求是我的应用程序应该只听到30秒后它应该抛出一个错误说“请求超时”。有什么建议吗?

        try
        {
            Byte[] bytes = new Byte[256];
            String data = null;
            // Enter the listening loop.
            while (true)
            {
                ReceiveTimer.Stop(); ;
                logger.Log("Waiting for a connection... ");

                // Perform a blocking call to accept requests.
                // You could also use server.AcceptSocket() here.
                TcpClient client = server.AcceptTcpClient();


                logger.Log("Connected!");

                data = null;

                // Get a stream object for reading and writing
                NetworkStream stream = client.GetStream();

                int i;

                // Loop to receive all the data sent by the client.
                while ((i = stream.Read(bytes, 0, bytes.Length)) != 0)
                {
                    // Translate data bytes to a ASCII string.
                    data = System.Text.Encoding.ASCII.GetString(bytes, 0, i);
                    logger.Log("Received: {0}", data);

                    // Process the data sent by the client.
                    data = data.ToUpper();

                    byte[] msg = System.Text.Encoding.ASCII.GetBytes(data);

                    // Send back a response.
                    stream.Write(msg, 0, msg.Length);
                    logger.Log("Sent: {0}", data);
                }

                // Shutdown and end connection
                client.Close();
                logger.Log("Shutdown and end connection");
            }
        }
        catch (SocketException ex)
        {
            logger.Log("SocketException: " + ex);
        }

2 个答案:

答案 0 :(得分:2)

我认为将ReceiveTimeout属性设置为30秒会为您处理。你试过了吗?

client.ReceiveTimeout = 30000;

该属性默认为0,因此您需要设置该属性。它抛出IOException。你可以抓住它并抛出你选择冒泡应用程序的例外。

答案 1 :(得分:0)

为什么要限制服务器监听的时间?

服务器应该一直在监听它正在运行。

无论如何,您可以创建一个正在侦听的任务(线程),并在超时后取消。