1

我有一个 WPF 服务器和客户端应用程序。当我启动服务器时，它开始收听传入的消息。但是，应用程序不能被触摸或关闭，它在监听中“卡住”了。我必须补充一点，它在处理消息等方面做了它应该做的事情。但我就是无法与表单交互。它与异步服务器套接字有关吗？我不确定要寻找什么...

这是我的服务器代码：

    private void startServer()
    {
        sck = new Socket(AddressFamily.InterNetwork, SocketType.Stream, ProtocolType.Tcp);
        sck.Bind(new IPEndPoint(0, serverPort));
        sck.Listen(100);

        while (true)
        {
            Socket accepted = sck.Accept();

            Buffer = new byte[accepted.SendBufferSize];
            int bytesRead = accepted.Receive(Buffer);

            byte[] formatted = new byte[bytesRead];

            for (int i = 0; i < bytesRead; i++)
            {
                formatted[i] = Buffer[i];
            }
            string command = Encoding.ASCII.GetString(formatted);
            string[] splittedCommand = command.Split(' ');

            jobsHistory.Items.Add(Encoding.ASCII.GetString(formatted));
            jobsHistory.Refresh();

            Process processToRun = new Process();
            processToRun.StartInfo.FileName = splittedCommand[0];
            processToRun.StartInfo.WorkingDirectory = Path.GetDirectoryName(splittedCommand[0]);
            processToRun.StartInfo.Arguments = "";
            for (int i = 1; i < splittedCommand.Length; i++)
            {
                processToRun.StartInfo.Arguments += " " + splittedCommand[i];
            }

            processToRun.Start();
            accepted.Close();
        }
    }
4

3 回答 3

1

如果这适用于 UI 线程，那么您将其捆绑在循环中。它没有机会处理任何 UI 事件。除此之外，Socket.Receive还有一个阻塞调用。

在本网站和 Google 上都有很多关于 BackgroundWorker` 类的文章。我建议你看看那些。

于 2013-01-30T08:09:25.317 回答
1

只需在单独的线程中执行监听，就像这样 -

    private void startServer()
    {
        sck = new Socket(AddressFamily.InterNetwork, SocketType.Stream, ProtocolType.Tcp);
        sck.Bind(new IPEndPoint(0, 8080));

        new Thread(() => StartListening(sck)).Start();
    }

    void StartListening(Socket socket)
    {
        socket.Listen(100);

        Accept(socket);
    }

    void Accept(Socket socket)
    {
        while (true)
        {
            Socket accepted = socket.Accept();

            Buffer = new byte[accepted.SendBufferSize];
            int bytesRead = accepted.Receive(Buffer);

            byte[] formatted = new byte[bytesRead];

            for (int i = 0; i < bytesRead; i++)
            {
                formatted[i] = Buffer[i];
            }
            string command = Encoding.ASCII.GetString(formatted);
            string[] splittedCommand = command.Split(' ');

            jobsHistory.Items.Add(Encoding.ASCII.GetString(formatted));
            jobsHistory.Refresh();

            Process processToRun = new Process();
            processToRun.StartInfo.FileName = splittedCommand[0];
            processToRun.StartInfo.WorkingDirectory = Path.GetDirectoryName(splittedCommand[0]);
            processToRun.StartInfo.Arguments = "";
            for (int i = 1; i < splittedCommand.Length; i++)
            {
                processToRun.StartInfo.Arguments += " " + splittedCommand[i];
            }

            processToRun.Start();
            accepted.Close();
        }

        // If you want to start listening again
        socket.Close();
        startServer();
    }
于 2013-01-30T08:09:51.987 回答
1

您应该使用BeginAccept在后台接受连接而不阻塞 UI 线程。当客户被接受时，BeginAccept再次调用 - 这会创建一个“循环”。然后你可以删除你的while循环。

该startServer方法可能如下所示：

private void startServer()
{
    sck = new Socket(AddressFamily.InterNetwork, SocketType.Stream, ProtocolType.Tcp);
    sck.Bind(new IPEndPoint(0, serverPort));
    sck.Listen(100);

    sck.BeginAccept(Accepted, sck);
}

然后，在Accepted方法中，执行以下操作：

private void Accepted(IAsyncResult result)
{
    Socket sck = result.AsyncState as Socket;
    Socket client = sck.EndAccept(result);

    sck.BeginAccept(Accepted, sck);
}


        Socket accepted = sck.Accept();
于 2013-01-30T08:20:43.490 回答