輸入錯誤後無法輸入提示

我有一個guessGameServerHandler類，它處理一個遊戲的單個實例爲客戶（遊戲是多線程）

我也有在輸入客戶端類讀數，那麼該處理程序響應。

我需要輸出一個錯誤，如果用戶輸入一個字符串，而不是一個整數，然後循環回來，再次要求輸入，但似乎無法得到它的工作。有什麼建議麼？

客戶端代碼：（這是while循環一做，檢查是否輸入等於正確的號碼或沒有）

// Read in clients input 
     System.out.print("Enter guess: "); 

      guess = Integer.parseInt(stdin.readLine()); 


     // send guess to server 
     out.write(String.format("%s%n", guess)); 
     out.flush();

guessGameHandlerServer：

do { 
      try{ 
      // Read in guesses from user 
      guess = Integer.parseInt(in.readLine()); 
      }catch(NumberFormatException e){ 
      } 

     gs.guess(guess); 
     String timeForm = decFor.format(gs.getTimeRemaining()/1000D); 
     // Check user input against target number & time left 
     if (gs.getTimeRemaining() < 1) { 
      System.out.println(ID + " - (LOSE)" + timeForm + "/" + gs.getGuesses()); 
      client.close(); 
     } else if (gs.getTimeRemaining() > 0 && guess > gs.getTarget() && guess < maxNum) { 
      out.write(String.format("HIGH:%s:%d%n", timeForm, gs.getGuesses())); 
      out.flush(); 
      System.out.println(ID + " " + guess + "(HIGH)-" + timeForm + "s/" + gs.getGuesses()); 
     } else if (gs.getTimeRemaining() > 0 && (guess < gs.getTarget() && guess > 0)) { 
      out.write(String.format("LOW:%s:%d%n", timeForm, gs.getGuesses())); 
      out.flush(); 
      System.out.println(ID + " " + guess + "(LOW)-" + timeForm + "s/" + gs.getGuesses()); 
     } else if (gs.getTimeRemaining() > 0 && guess == gs.getTarget()) { 
      out.write(String.format("WIN:%d%n", gs.getGuesses())); 
      out.flush(); 
      System.out.println(ID + " " + guess + "(WIN)-" + timeForm + "s/" + gs.getGuesses()); 
     } else if (gs.getTimeRemaining() > 0 && (guess >= maxNum || guess < 1)) { 
      out.write(String.format("ERR:%d%n", gs.getGuesses())); 
      out.flush(); 
      System.out.println(ID + " ** (ERR out of range)" + timeForm + "/" + gs.getGuesses()); 
     } 


     } while (true);

與當前的代碼，輸入字符串只是使客戶端線程關閉，不知道爲什麼。

來源

2016-04-25 Travisty

在這樣的客戶端處理異常：

while(true) { 
    try { 
     System.out.print("Enter guess: "); 

     guess = Integer.parseInt(stdin.readLine()); 
     break; 
    } 
    catch(Exception ex) { 
      System.out.println("Try again"); 
    } 
} 
    // send guess to server 
    out.write(String.format("%s%n", guess)); 
    out.flush();

來源

2016-04-25 22:55:21 AhmadWabbi

這有效，但我也試圖發送一條消息到服務器，當客戶端輸入一個不正確的輸入。將'out.write'放在catch塊中似乎不起作用。 – Travisty

當用戶輸入錯誤的輸入時，你想發送給服務器的是什麼？您無法確定地發送「猜測」，因爲由於用戶輸入不正確而導致它沒有值。無論你想在這種情況下發送什麼，把它放在「catch」塊，然後是「flush」。它應該工作。 – AhmadWabbi

你需要有一個try catch塊中做while循環趕上NumberFormatException異常。

try{ 
     System.out.print("Enter guess: "); 

     guess = Integer.parseInt(stdin.readLine()); 
    }catch(NumberFormatException ne){ 
    //exception handling - set the do while loop condition to true so that it is called again 
    } 


    // send guess to server 
    out.write(String.format("%s%n", guess)); 
    out.flush();

來源

2016-04-25 22:56:40 raag

輸入錯誤後無法輸入提示

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