Question

我目前正在为大学建立一个Java程序。部分原因是建立一个菜单系统。导师希望在菜单上进行异常处理。这是一个没有用户界面的简单菜单。

菜单是一个switch语句，用户输入他们想要的选项;

选项一
选项二
选项三

导师想要的是一个try catch方法，一个用户可以输入数字1或输入“One”，系统不会崩溃。

我在网上看过无数篇文章，但我要么找不到这方面的帮助，要么就是我的头脑。任何帮助都会很棒。

编辑：所以这是第一年的课程，他只是希望功能下降，我们可以为他做好准备。我现在可以得到代码。首先是PremiumAccount类，然后主菜单在程序类中调用每个方法。您可以在这里找到下面的菜单，如果需要，我也可以提供PremiumAccount课程。

做{

        System.out.println("Premium Phone!");
        System.out.println("");
        System.out.println("Please select the option for which you would like to do!");
        System.out.println("1. Make a Call");
        System.out.println("2. Send a Text");
        System.out.println("3. Top Up");
        System.out.println("4. Display your details");
        System.out.println("5. Get your balance");
        System.out.println("6. Get your minutes used");
        System.out.println("7. Get your texts sent");
        System.out.println("8. Update the Call Costs");
        System.out.println("9. Update the Text Costs");
        System.out.println("10. Update the discount rate");
        System.out.println("11. Exit");
        intMenuChoice = numInput.nextInt(); //user input menu choice

        switch(intMenuChoice)
        {
            case 1: System.out.print("Please enter the amount length of the call in minutes: ");
                    intMinutesUsed = numInput.nextInt();
                    System.out.println(pre123.makeCall(intMinutesUsed));
                    System.out.println("");
                break;
            case 2: System.out.print("Please enter the amount of characters used in the text: "); 
                    intCharactersUsed = numInput.nextInt();
                    System.out.println(pre123.sendText(intCharactersUsed));
                    System.out.println("");
                break;
            case 3: System.out.print("Please enter the amount you would like to Top Up: ");
                    dblTopUp = numInput.nextDouble();
                    System.out.println(pre123.topUp(dblTopUp));
                    System.out.println("");
                break;
            case 4: System.out.println("***Account Details***");
                    pre123.displayAccountDetails();
                    System.out.println("");
                break;
            case 5: System.out.println("Account Balance: £" +df.format(pre123.getBalance()));
                    System.out.println("");
                break;
            case 6: System.out.println("Minutes Used: " +pre123.getMinutes());
                    System.out.println("");
                break;
            case 7: System.out.println("Texts Used: " +pre123.getTexts());
                    System.out.println("");
                break;
            case 8: System.out.print("Please enter the new Cost of a Call: ");
                    newCallCost = numInput.nextDouble();
                    System.out.println(pre123.updateCallCost(newCallCost));
                    System.out.println("");
                break;
            case 9: System.out.println("Please enter the next Cost of a Text: ");
                    newTextCost = numInput.nextDouble();
                    System.out.println(pre123.updateTextCost(newTextCost));
                    System.out.println("");
                break;
            case 10: System.out.println("Please enter the new rate of Discount");
                     newDiscount = numInput.nextDouble();
                     System.out.println(pre123.setNewDiscount(newDiscount));
                     System.out.println("");
                break;
            case 11: System.out.println("Have a nice Day!");
                     menuStatus = 1;
                break;

            default: System.out.println("Please select a number between 1 - 11");

        }//end switch statement


    }while(menuStatus == 0);//end do while

我们只需要按照我的说法让用户能够输入他们想要的但是尝试捕获，这样他们就可以输入“1”或“一个”，如果他们键入它，系统不会崩溃< / p>

提前多多感谢康纳

Answer 1

您可以使用Integer.parseInt()并抓住NumberFormatException，然后检查String

编辑后

您可以使用intMenuChoice = numInput.nextInt();代替String input = numInput.nextLine();，然后添加以下内容

try{
   intMenuChoice = Integer.parseInt(input);
}catch (NumberFormatException e){
   switch(input.toLowerCase()){
      case "one":
         intMenuChoice = 1;
         break;
      ......//So on

   }
}

Answer 2

public static void main(String[] args) {
    Scanner scan = new Scanner(System.in);
    String input = "";

    System.out.print("Enter the menu item: "); 
    input = scan.NextLine();

    try{
        if(input.equals("One") || input.equals("1"))
            doOptionOne();
        else
            throw new Exception("The menu option is not 1 or One"); //throws exception if the choice is not valid 
    }
    catch(Exception e){
        System.out.println(e.getMessage()); //print the exception message
    }
}

菜单系统的预期处理

2 个答案: