Java 在用户输入不正确时使用JOptionPane循环

我正在用java模拟一个彩票游戏。现在一切正常，我只验证用户输入数字1到100>我还想阻止用户输入和清空值。现在正在使用try。。接住但这只是第一次起作用。用户输入6个数字。假设第一个输入为空，则会显示错误，但如果用户在空输入上再次按enter键，程序将崩溃。我无法让它循环，我尝试了好几件事，但都没有成功。下面是获取用户输入的代码

String [] charNums = {"1st","2nd","3rd","4th","5th","Bonus"};
//get input 5 numbers from user 
for(int i=0; i<6;i++){

    //boolean correctInput = false;
    //while(!correctInput){
         try {
            inputNumbers[i] = Integer.parseInt(JOptionPane.showInputDialog("Enter "+charNums[i]+" number from 1 to 100"));
                while(!validate(inputNumbers[i])){
                     JOptionPane.showMessageDialog(null,"Invalid Number! try Again","ERROR",JOptionPane.ERROR_MESSAGE);
                     inputNumbers[i] = Integer.parseInt(JOptionPane.showInputDialog("Enter "+charNums[i]+" number from 1 to 100"));
                }
                ////check for duplicate entries from user
                for (int k=0; k<i; k++)  {    
                    while (k!=i && inputNumbers[k] == inputNumbers[i])  {
                        JOptionPane.showMessageDialog(null,"Duplicate Entry! try again","ERROR",JOptionPane.ERROR_MESSAGE);
                        inputNumbers[i] = Integer.parseInt(JOptionPane.showInputDialog("Enter "+charNums[i]+" number from 1 to 100"));
                    }
                }
                //correctInput = true;
                //break;
            }catch (NumberFormatException e) {
                JOptionPane.showMessageDialog(null,"Number not entered! try Again","ERROR",JOptionPane.ERROR_MESSAGE);
                //throw new NumberFormatException("not number");
                inputNumbers[i] = Integer.parseInt(JOptionPane.showInputDialog("Enter "+charNums[i]+" number from 1 to 100"));
                //correctInput = false;
                //JOptionPane.setValue(JOptionPane.UNINITIALIZED_VALUE);
                //continue;

            //}
        }
        userNumbers[i] = inputNumbers[i];
    }

试一下

 boolean correctInput = false;
        ///create array to display user
        String [] charNums = {"1st","2nd","3rd","4th","5th","Bonus"};
        //get input 5 numbers from user 
        for(int i=0; i<6;i++){
            do {
                try {
                    inputNumbers[i] = Integer.parseInt(JOptionPane.showInputDialog("Enter "+charNums[i]+" number from 1 to 100"));
                    while(!validate(inputNumbers[i])){
                        JOptionPane.showMessageDialog(null,"Invalid Number! try Again","ERROR",JOptionPane.ERROR_MESSAGE);
                        inputNumbers[i] = Integer.parseInt(JOptionPane.showInputDialog("Enter "+charNums[i]+" number from 1 to 100"));
                    }

                    ////check for duplicate entries from user
                    for (int k=0; k<i; k++)  {    
                        while (k!=i && inputNumbers[k] == inputNumbers[i])  {
                            JOptionPane.showMessageDialog(null,"Duplicate Entry! try again","ERROR",JOptionPane.ERROR_MESSAGE);
                            inputNumbers[i] = Integer.parseInt(JOptionPane.showInputDialog("Enter "+charNums[i]+" number from 1 to 100"));
                        }
                    }
                    correctInput = true;
                    //break;
                }catch (NumberFormatException e) {
                    JOptionPane.showMessageDialog(null,"Number not entered! try Again","ERROR",JOptionPane.ERROR_MESSAGE);
                    //throw new NumberFormatException("not number");
                    inputNumbers[i] = Integer.parseInt(JOptionPane.showInputDialog("Enter "+charNums[i]+" number from 1 to 100"));
                    correctInput = false;
                    //JOptionPane.setValue(JOptionPane.UNINITIALIZED_VALUE);
                    //continue;

                }
            }while(!correctInput);
            userNumbers[i] = inputNumbers[i];
        }

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try/catch应该在while循环中，否则在抛出异常时退出循环

---

提示：do/while。

这里是您的代码修改，使循环6个数字变得更加简单，我已经测试过了，它运行良好。如果有什么不清楚，请告诉我

   String [] charNums = {"1st","2nd","3rd","4th","5th","Bonus"};
   int[] inputNumbers = new int[6];
   //get input 6 numbers from user 
   int cnt = 0;
   while(true)
   {
     try
     {
       String tmp = JOptionPane.showInputDialog("Enter "+charNums[cnt]+" number from 1 to 100");
       int val = Integer.parseInt(tmp);
       boolean duplicate = false;
       for (int k=0; k<cnt; k++)
       {
         if(val==inputNumbers[k])
         {
           JOptionPane.showMessageDialog(null,"Duplicate Entry! try again","ERROR",JOptionPane.ERROR_MESSAGE);
           duplicate = true;
           break;
         }   
       }   
       if(duplicate)continue;

       inputNumbers[cnt] = val;
       cnt++;
     }
     catch(Exception e)
     {
     }
     if(cnt==6)break;
   }

---

上面的代码对我不起作用。但是我通过将try-and-catch放在validate方法中进行了修复，现在每当有错误的输入时，它都会循环。这是密码

public static boolean validate(String num){

    try {
        int convertedNum = Integer.parseInt(num);
        if(convertedNum < 0 || convertedNum > 100){
            return false;
        }else{
            return true;
        }
    }catch ( NumberFormatException e){
        return false;
    }
}

for(int i=0; i<6;i++){

            inputNumbers[i] = JOptionPane.showInputDialog("Enter "+charNums[i]+" number from 1 to 100");
            while(!validate(inputNumbers[i])){
                if(inputNumbers[i] == null){
                    int stopGame = JOptionPane.showConfirmDialog(null,"Do you wish to cancel the game? All progress will be lost","",JOptionPane.YES_NO_OPTION);
                    if(stopGame == JOptionPane.YES_OPTION){
                        System.exit(0);
                    }
                }
                JOptionPane.showMessageDialog(null,"Invalid Number! try Again","ERROR",JOptionPane.ERROR_MESSAGE);
                inputNumbers[i] = JOptionPane.showInputDialog("Enter "+charNums[i]+" number from 1 to 100");
            }

            ////check for duplicate entries from user
            for (int k=0; k<i; k++)  {    
                while (k!=i && Integer.parseInt(inputNumbers[k]) == Integer.parseInt(inputNumbers[i]))  {
                    JOptionPane.showMessageDialog(null,"Duplicate Entry! try again","ERROR",JOptionPane.ERROR_MESSAGE);
                    inputNumbers[i] = JOptionPane.showInputDialog("Enter "+charNums[i]+" number from 1 to 100");
                }
            }

            //convert string to int
            userNumbers[i] = Integer.parseInt(inputNumbers[i]);
        }

---

你是说在里面！验证循环？是的，这就是我的意思。想一想，这很简单。我可以给你解决方案，但如果你自己解决它，你将学会更多如何思考编程，并感到更有成就感；请看我在答案中的暗示。我知道我已经做了一整天了，也许你能解释一下“捉迷藏”是如何起作用的？我会了解更多的。我将发布我现在所拥有的。Netbeans告诉我，correctInput从未被使用过，如果我了解try-and-catch的工作原理，我可能会想出答案