如何通過Java套接字從客戶端向服務器發送多個映像？

Question

更新

我正在嘗試從客戶端向服務器發送多個圖像。 客戶端有兩個線程

1. 線程1正在截屏

2. 線程2正在向服務器發送屏幕截圖

這段代碼正在拍2張截圖。 但只有第一個屏幕截圖成功保存在服務器上。 請幫我發送並保存多個圖像到服務器。

沒有錯誤，沒有例外。

輸出：

thread1正在運行...

thread2正在運行...

thread1正在運行...

thread2正在運行...

客戶

abstract class ScreenCapture implements Runnable{

 static BufferedImage screencapture;
 static ByteArrayOutputStream baos;
 static byte[] ImageInBytes;

   public static void main(String args[]) throws
       AWTException, IOException, InterruptedException {

  // Open your connection to a server, at port 1234
  final Socket ClientSocket = new Socket("localhost",1234);

  final DataOutputStream dos= new DataOutputStream(ClientSocket.getOutputStream());
  DataInputStream in = new DataInputStream(ClientSocket.getInputStream() );
  baos = new ByteArrayOutputStream();

  try{
      //First thread that is Taking screenshot
      Thread TakeScreenShotthread = new Thread () 
      {
          public void run () {        

          // Capture Screen using BufferedImage Library
           try {
               screencapture = new Robot().createScreenCapture(
               new Rectangle(Toolkit.getDefaultToolkit().getScreenSize()) );
               System.out.println("thread1 is running...");

            } catch (HeadlessException e) {

                    e.printStackTrace();
            } catch (AWTException e) {

                    e.printStackTrace();
            }
        } 
    };

    //Thread 2 that is Sending Screenshot to server
    Thread sendingScreenShotThread =new Thread () {
          public void run () {
              //Sending Screen to Server
               try {
                      ImageIO.write(screencapture, "jpg", baos);
                      ImageInBytes = baos.toByteArray();
                      dos.write(ImageInBytes);
                     // File Rif = new File(System.currentTimeMillis() + ".jpg");
                      //ImageIO.write(screencapture, "jpg", Rif);
                      System.out.println("thread2 is running...");

                } catch (IOException e) {
                e.printStackTrace();
                }       
               finally{

                    try {
                        baos.flush();
                    } catch (IOException e) {
                        // TODO Auto-generated catch block
                        e.printStackTrace();
                    }
                }          
          }
        };
        TakeScreenShotthread.start();
        TakeScreenShotthread.sleep(1000);
        sendingScreenShotThread.start();
        sendingScreenShotThread.sleep(1000);
        TakeScreenShotthread.run();
        sendingScreenShotThread.run();
  }finally
  {
       //Closing Clients
                in.close();
                baos.close();
                ClientSocket.close();
  }  
  }
 }

服務器

    public class ServerConnection
    { 
    public static void main(String args[]) throws IOException
    { 
    ServerSocket serversock = new ServerSocket(1234);
    Socket clientSocket = null;
    BufferedReader BF_RecievingGUID;
    clientSocket = serversock.accept();
    InputStream in=clientSocket.getInputStream();
    OutputStream out = null; 

   try{    
        boolean processing=true;
       while(processing)
      {
       try {
        byte[] buffer = new byte[1024];
        out = new BufferedOutputStream(new FileOutputStream(path));

       while ((in.read(buffer)) >= 0) { 
             out.write(buffer);
       }
       System.out.println("Image file written successfully");
   } catch (Exception e) {
   }finally {
       processing=false;
        if (out != null) out.close();
     }
   }
  }
  finally{
        BF_RecievingGUID.close();
        out.close();
      clientSocket.close();
        serversock.close();

        }   
    }
  }

Answer 1

它只會讀取一次，因為你打破了while循環：

if(RecievedImage != null)
{
     ImageIO.write(RecievedImage, "jpg", RecievedImageFile);
     RecievedImage.flush();
     System.out.println("Image file written successfully");
}else{
     System.out.println("image is empty");
}break;  //you break here 

我不明白為什么你這樣做：

new Thread(new ThreadHandler(clientSocket)).start();