為什么我的 try catch 方法沒有顯示我為輸入的字符串而不是 integer 編寫的錯誤消息？ （爪哇）

Question

System.out.println(" ");
System.out.println("Stage 4: Enter an integer between 5 and 500:");
IntEntry = keyedInput.nextInt();

if(IntEntry > 5 && IntEntry < 500)
{
    System.out.println("You will go to stage 5.");
}
else
{
    try
    {
        while(IntEntry < 5 || IntEntry > 500)
        {
            System.out.println("Try Again. Enter an integer that is between 5 and 500 characters long. “z”.");
            IntEntry = keyedInput.nextInt();
        }
    }
    catch(NumberFormatException e)
    {
        System.out.println("Try Again. Enter an integer between 5 and 500:.");
    }
}

在上面的代碼中，我試圖獲取用戶在 5 到 500 之間輸入的有效 integer。如果用戶輸入的是字符串而不是integer ，程序應該使用try-catch方法向用戶發送一條消息 output。 try-catch方法對我不起作用，它會輸出一條錯誤消息。 有人能告訴我為什么try-catch方法在我的代碼中不起作用嗎？

Answer 1

Scanner 的nextInt方法的 javadoc聲明它拋出InputMismatchException 。 不是NumberFormatException 。 （您對Integer.parseInt感到困惑）。

Answer 2

像這樣更新你的代碼

System.out.println(" ");
System.out.println("Stage 4: Enter an integer between 5 and 500:");

try{        
     IntEntry = keyedInput.nextInt();
     if(IntEntry > 5 && IntEntry < 500)
        {
            System.out.println("You will go to stage 5.");
        }
     else
        {
            try
            {
                while(IntEntry < 5 || IntEntry > 500)
                {
                    System.out.println("Try Again. Enter a string that is between 5 and 15 characters long, and does not contain the letter “z”.");
                    IntEntry = keyedInput.nextInt();
                }
            }catch(NumberFormatException e)
            {
                System.out.println("Try Again. Enter a string that is between 5 and 15 characters long, and does not contain the letter “z”.");
            }
    }
}catch(NumberFormatException e)
                {
                    System.out.println("Try Again. Enter a string that is between 5 and 15 characters long, and does not contain the letter “z”.");
                }    
          

  

Answer 3

使用Scanner#nextInt()時，您應該捕獲InputMismatchException而不是NumberFormatException 。 此外，您第一次使用nextInt()不在try-catch塊內。 此外，當您稍后捕獲異常時，您的代碼不再循環。 您可以使用一個do-while循環和一個標志（例如valid ）來干凈地執行此操作。 只有當 integer 可以被解析並且在范圍內時，才會設置標志。 因此，循環繼續，直到用戶輸入范圍內的有效 integer。 這是一個完整的例子：

Scanner scanner = new Scanner(System.in);

System.out.println("Stage 4: Enter an integer between 5 and 500:");

int intEntry;
boolean valid = false;
do {
    try {
        intEntry = scanner.nextInt();
        if (intEntry > 5 && intEntry < 500) {
            valid = true; // set flag to leave loop
        } else {
            System.out.println("Integer out of range. Try Again. Enter an integer between 5 and 500:");
        }
    } catch (InputMismatchException e) {
        System.out.println("Could not parse integer. Try again. Enter an integer between 5 and 500:");
        scanner.next(); // clear invalid token
    }
} while (!valid);

System.out.println("You will go to stage 5.");