Java猜測遊戲服務器 - 客戶端

Question

我正在做猜謎遊戲。所有工作正常，但我有一個問題。第一次，我要求一個數字它正確地說我是否更高或更低，但第二沒有說什麼都沒有..

這是我的服務器和客戶端代碼：

public class ServerNum 
    { 
    public static void main(String[] args) throws IOException 
    { 
     adivinarNum juego = new adivinarNum(); 
     ServerSocket socket = null; 
     Socket client = null; 
     String resultado; 
     boolean correcto = false; 
     int intentos; 

       try{ 
        socket = new ServerSocket(1234); 

       }catch(IOException ioe) 

       { 
        System.err.println(ioe); 
        return; 
       } 
        System.out.println("El servidor sigue funcionando..."); 
        client = socket.accept(); 
        System.out.println("El cliente se ha conectado"); 
        DataInputStream in = new DataInputStream(new BufferedInputStream(client.getInputStream())); 
        DataOutputStream out = new DataOutputStream(new BufferedOutputStream(client.getOutputStream())); 
     while(!correcto) 
      { 
       intentos = in.readInt(); 
       resultado = juego.adivinar(intentos); 
       correcto = juego.getCorrecto(); 
       out.writeUTF(resultado); 
       out.writeBoolean(correcto); 
       out.flush(); 
      if(correcto == false){ 
       client = socket.accept(); 
       intentos = in.readInt(); 
       resultado = juego.adivinar(intentos); 
       correcto = juego.getCorrecto(); 
       out.writeUTF(resultado); 
       out.writeBoolean(correcto); 
       out.flush(); 
       } 
      else{ 
       client.close(); 
       socket.close(); 
       } 

      } 
     } 
    } 


     public class ClientNum 
     { 
     public static void main(String[] args) throws IOException 
     { 
      System.out.println("This is Number Guessing Game. \nChoose any number between 1 to 1000 : "); 
      Scanner keyboard = new Scanner(System.in); 
      int attempt = 0; 
      try 
      { 
      attempt = keyboard.nextInt(); 
      if(attempt < 1 || attempt > 999) 
      { 
       System.out.println("Your number is too large/small, please make a guess between 1 to 1000"); 
       attempt = keyboard.nextInt(); 
       } 
      } 
     catch(NumberFormatException nfe) 
     { 
      System.out.println("Just choose numbers! Try again"); 
      attempt = keyboard.nextInt(); 
     } 

     try 
       { 
        Socket server = new Socket("localhost", 1234); 
        System.out.println("Connecting..."); 

        DataOutputStream out = new DataOutputStream(new BufferedOutputStream(server.getOutputStream())); 
        DataInputStream in = new DataInputStream(new BufferedInputStream(server.getInputStream())); 

        out.writeInt(attempt); 
        out.flush(); 
        System.out.println("Our server is still running..."); 
        String result = in.readUTF(); 
        boolean correct = in.readBoolean(); 
        System.out.println(result); 


        while (!correct){ 
          attempt = keyboard.nextInt(); 
          out.writeInt(attempt); 
          out.flush(); 
          System.out.println("Our server is still running..."); 
          result = in.readUTF(); 
          System.out.println(result); 
          correct = in.readBoolean(); 
        } 

         server.close(); 
         System.out.println("Finish. Thank you"); 
         System.exit(0); 

       }catch(IOException ioe){ 
        System.err.println(ioe); 
       } 
      } 
     } 

我會感謝任何幫助！謝謝！

Answer 1

socket.accept用於接受到客戶端的連接。這意味着它用於開始與客戶端通信，因此，每個連接只需要調用一次。此外，您的客戶在獲得正確答案後立即刪除連接，然後到達其方法main()的末尾，這意味着該程序在那裏結束。

你應該設計你的程序，以便建立連接，然後在while循環內（通常稱爲「遊戲循環」）播放遊戲代碼，直到用戶不想再玩。當用戶不想再玩時，允許循環結束，然後讓它關閉與服務器的連接，以便應用程序可以正常關閉。

應該指出，accept方法有一個效率開銷與它相關聯，因此如果你不必經常調用它，程序將更有效地運行。