在一次接收频繁请求时获得套接字超时

Question

我试图用我的手机为我的笔记本电脑构建一个遥控器。我编写了一个服务器应用程序，在我的笔记本电脑上运行，远程应用程序用作服务器应用程序的客户端。我想实现一个鼠标垫，问题是当我在"触摸板"上移动手指太快时，我在几次迭代后在服务器端接收读取超时。

服务器代码final ExecutorService clientProcessingPool = Executors.newFixedThreadPool(20);

    Runnable serverTask = new Runnable()
    {
        @Override
        public void run()
        {
            ServerSocket serverSocket = null;
            try
            {
                serverSocket = new ServerSocket(DEFAULT_PORT);
                serverSocket.setReuseAddress(true);
                _working = true;
                while (_working)
                {
                    Socket clientSocket = serverSocket.accept();
                    clientSocket.setSoTimeout(10000);
                    clientProcessingPool.submit(new ClientTask(clientSocket));
                }
            }
            catch (IOException e)
            {
                System.err.println("Unable to process client request");
                e.printStackTrace();
            }
            finally
            {
                try
                {
                    if (serverSocket != null)
                    {
                        serverSocket.close();
                    }
                }
                catch (IOException e)
                {
                    e.printStackTrace();
                }
            }
        }
    };
    Thread serverThread = new Thread(serverTask);
    serverThread.start();
private class ClientTask implements Runnable
{
    private final Socket    clientSocket;
    private ClientTask(Socket clientSocket)
    {
        this.clientSocket = clientSocket;
    }
    @Override
    public void run()
    {
        System.out.println("Got a client !");
        try
        {
            System.out.println("Connected!");
            DataOutputStream dOut = new DataOutputStream(clientSocket.getOutputStream());
            DataInputStream dIn = new DataInputStream(clientSocket.getInputStream());
            String request = dIn.readUTF();
            parseRequest(request);
            System.out.println("request=" + request);
            dOut.writeUTF("Got the command");
            dOut.flush(); // Send off the data
            dIn.close();
            dOut.close();
            clientSocket.close();
        }
        catch (Exception e)
        {
            e.printStackTrace();
        }
    }

我想也许只有当从起始位置到结束位置的距离大于STEPS(一个常数)时，我才会发送请求，然后才发送请求。但我认为鼠标移动不流畅。

谢谢。

Answer 1

RXAndroid和RXJava非常适合异步处理这些请求，并允许您为频繁的请求添加一个debounce。查看react .io