尝试在 do while 循环中捕获

Question

程序要求用户输入 double num 1 和 double num 2 如果有异常，我希望它再次要求输入 num 1 和 num 2

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);
}

然后，当我尝试代码时，它是否会通过在 num1 或 num 2 中输入字符串来捕获异常，我有这个无限循环： 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

你需要在循环内重置error变量

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

Answer 2

没有必要使用异常处理。只需使用Scanner.hasNextDouble()方法判断实际用户输入是否为double，否则继续循环。

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 3

它在 C# 中，但比较相似:)

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

对于实际代码（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 4

您可能想测试是否没有错误：

}while(error != 1);

或

}while(error == 0);

Answer 5

如果输入无效，您将需要一个调用自身的输入方法。

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);
    }    
}

并在您的其他方法中调用此方法两次。

Answer 6

它可能看起来很难看，但这是一种方法

  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 7

您需要在catch 块内添加sc.next();。

nextDouble 方法在出现异常时不会清除缓冲区。所以下次你调用它时你会得到同样的错误，因为旧的输入仍然在缓冲区中。

您还需要在循环开始时重置 error 标志。

Answer 8

您必须将 sc.next(); 放在 catch 中，这样它才能清除您的扫描仪变量并要求输入