嘗試在do while while循環中捕獲

Question

程序要求用戶輸入雙數字1和雙數字2 ，如果有例外，我希望它再次詢問數字1和數字2的輸入

public static void main (String[] args) { 
    Scanner sc = new Scanner(System.in); 
    double num1, num2; 
    int error = 0; 
    int text; 
    System.out.print("Enter 4 "); 
     text = sc.nextInt(); 

    do{ 

     try{ 

    if(text == 4){ 
      System.out.print("Enter number 1: "); 
      num1 = sc.nextDouble(); 
      System.out.print("Enter number 2: "); 
      num2 = sc.nextDouble(); 
      double quotient = num1/num2; 
      System.out.println("The Quotient of "+num1 + "/" +num2+ " = "+quotient); 
      } 
     }catch(Exception ex){ 
      System.out.println("You've entered wrong input"); 
       error = 1; 
     }         

      }while(error == 1); 
} 

然後當我嘗試代碼，如果它會通過在NUM1或num輸入查詢字符串捕獲異常2我在這個無限循環： Enter number 1: You've entered wrong input Enter number 1: You've entered wrong input Enter number 1: You've entered wrong input Enter number 1: You've entered wrong input Enter number 1: You've entered wrong input

Answer 1

您需要的error變量重新循環內

do { 
    error = 0; 
    //... 
} while(error == 1); 

Answer 2

你可能想測試，如果沒有錯誤：

}while(error != 1); 

或

}while(error == 0); 

Answer 3

你所需要的自稱，如果輸入的是無效的輸入方法。

double getInput(Scanner sc) { 
    try { 
     double num = sc.nextDouble(); 
     return num; 
    } catch(Exception ex) { 
     System.out.println("You've entered wrong input"); 
     return getInput(sc); 
    }  
} 

並用另一種方法調用這個方法兩次。

Answer 4

它可能看起來很醜陋，但這裏是一個辦法做到這一點

do 
    { 
    if(...) 
    { 

     boolean successReading = false; 
     while(!successReading) 
     { 
     try 
     { 
      System.out.print("Enter number 1: "); 
      num1 = sc.nextDouble(); 
      System.out.print("Enter number 2: "); 
      num2 = sc.nextDouble(); 

      successReading = true; 

      double product = num1*num2; 
     } 
     catch(Exception e) 
     { 
      successReading = false; 
     } 
     } 

    } 
    }while(...) 

Answer 5

它在C＃中，但比較相似:)

public class Program 
{ 
    private static double ReadUserInput (string message) 
    { 
    // This is a double 
    // The '?' makes it nullable which is easier to work with 
    double? input = null; 

    do 
    { 
     // Write message out 
     Console.Write(message); 
     // Read answer 
     var inputString = Console.ReadLine(); 
     // Temp variable for the number 
     double outputNumber = 0; 
     // Try parse the number 
     if (double.TryParse(inputString, out outputNumber)) 
     { 
     // The number was parsable as a double so lets set the input variable 
     input = outputNumber; 
     } 
     else 
     { 
     // Tell the user the number was invalid 
     Console.WriteLine("Sorry bud, but '" + inputString + "' is not a valid double"); 
     } 
    } 
    while (input == null); // Keep running until the input variable is actually set by the above 

    // Return the output 
    return (double)input; 
    } 

    public static void Main() 
    { 
    // Read a number 
    var num1 = ReadUserInput("Enter number 1:"); 
    // Read another number 
    var num2 = ReadUserInput("Enter number 2:"); 
    // Show the calculation 
    Console.WriteLine("Answer: " + (num1*num2)); 
    } 
} 

Demo

而對於實際的代碼（在JAVA中）：

public class JavaFiddle 
{ 
    public static void main (String[] args) 
    { 
    // Read a number 
    Double num1 = ReadUserInput("Enter number 1:"); 
    // Read another number 
    Double num2 = ReadUserInput("Enter number 2:"); 
    // Show the calculation 
    System.out.println("Answer: " + (num1*num2)); 
    } 

    public static Double ReadUserInput (String message) 
    { 
    java.util.Scanner inputScanner = new java.util.Scanner(System.in); 
    Double input = null; 

    do 
    { 
     // Write message out 
     System.out.println(message); 
     // Read answer 
     String inputString = inputScanner.nextLine(); 
     try 
     { 
     // Try parse the number 
     input = Double.parseDouble(inputString); 
     } 
     catch (NumberFormatException e) 
     { 
     // Tell the user the number was invalid 
     System.out.println("Sorry bud, but '" + inputString + "' is not a valid double"); 
     } 
    } 
    while (input == null); // Keep running until the input variable is actually set by the above 

    // Return the output 
    return input; 
    } 
} 

Answer 6

沒有必要利用異常處理。只需使用Scanner.hasNextDouble()方法來確定實際用戶輸入是否爲雙倍，否則繼續循環。

package com.company; 

import java.util.Scanner; 

public class Main { 

    public static void main(String[] args) { 
     Scanner sc = new Scanner(System.in); 
     double num1, num2; 
     num1 = readDouble(1, sc); 
     num2 = readDouble(2, sc); 
     double quotient = num1/num2; 
     System.out.println("The Quotient of " + num1 + "/" + num2 + " = " + quotient); 
    } 

    private static double readDouble(int i, Scanner sc) { 
     while (true) { 
      System.out.print("Enter number " + i + ": "); 
      if (!sc.hasNextDouble()) { 
       System.out.println("You've entered wrong input"); 
       sc.next(); 
       continue; 
      } 
      break; 
     } 
     return sc.nextDouble(); 
    } 
} 

Answer 7

您需要添加sc.next();內catch塊。

nextDouble如果發生異常，方法不會清除緩衝區。所以下次你調用它時，你會得到相同的錯誤，因爲舊的輸入仍然在緩衝區中。

此外，您還需要在循環開始時重置error標誌。