在JAVA中使用Try / Catch

Question

我有以下代码试图让用户仅输入整数； 如果输入了字符串，则会显示系统错误消息“请仅输入数字”，然后再次显示“输入您的ID＃：”。 我尝试使用try / catch方法，但未正确使用它-仍然是初学者。 我知道我可以使用“ NumberFormatException ”，但不确定在哪里。 有人可以帮忙吗？ 谢谢！

        //Get Customer ID and Account Number
        do
        {   System.out.print("Enter your ID#: ");
            custid = Integer.parseInt(input.readLine());
            System.out.print("Enter your Account Number#: ");
            custacctnum = Integer.parseInt(input.readLine());
        //validate the choice
            for(int i=0; i<people.length; i++)  
            {   if ((people[i].custid == custid) && (people[i].custacctnum == custacctnum))
                {   match = true;
                System.out.println("Welcome "  +people[i].firstname+ " to JJ Dealership!");
                for(int p=0; p<cluster.length; p++)
                    System.out.println(+(p+1)+ ": " +cluster[p].year+"," +cluster[p].make+ "," +cluster[p].model);

                System.out.println(people[i].firstname+ ", what color car would you like?");
                    break;
                }
            }
            if (!match)
            {   System.out.println("Invalid ID");   
        } while (!(match));

Answer 1

例外情况应在特殊情况下进行。 我建议您使用Scanner和hasNextInt()仅在它不是int时continue 。 那就是让Scanner像

Scanner input = new Scanner(System.in);

然后这样的事情会起作用，

do {
    System.out.print("Enter your ID#: ");
    if (!input.hasNextInt()) {
        System.out.printf("%s is not an int.%n", input.nextLine());
        continue;
    }
    custid = input.nextInt();
    System.out.print("Enter your Account Number#: ");
    custacctnum = input.nextInt();
    if (!input.hasNextInt()) {
        System.out.printf("%s is not an int.%n", input.nextLine());
        continue;
    }

如果您真的想使用try-catch它应该看起来像这样，

do
{
    try {
        System.out.print("Enter your ID#: ");
        custid = Integer.parseInt(input.readLine());
        System.out.print("Enter your Account Number#: ");
        custacctnum = Integer.parseInt(input.readLine());
    //validate the choice
        for(int i=0; i<people.length; i++)  
        {   if ((people[i].custid == custid) &&
                    (people[i].custacctnum == custacctnum))
            {   match = true;
            System.out.println("Welcome "  +people[i].firstname
                    + " to JJ Dealership!");
            for(int p=0; p<cluster.length; p++)
                System.out.println("" + (p+1) + ": " +cluster[p].year+","
                        + cluster[p].make+ "," +cluster[p].model);

            System.out.println(people[i].firstname
                     + ", what color car would you like?");
                break;
            }
        }
    } catch (NumberFormatException nfe) {
        nfe.printStackTrace();
    }
} while (!(match));

Answer 2

@Elliott Frisch，在看到您的答案并弄清楚如何使用try / catch之前，我已经对其进行了“玩耍”。 这很简单，但是我将看看您发布的另一个选项。 当用户输入字符串时，使用下面的更新代码，它将显示错误消息，然后将其带到“输入您的ID＃”行以重试。

非常感谢大家的迅速答复。

do
        {   
            try{
                System.out.print("Enter your ID#: ");
                custid = Integer.parseInt(input.readLine());
                System.out.print("Enter your Account Number#: ");
                custacctnum = Integer.parseInt(input.readLine());
        //validate the choice
            for(int i=0; i<people.length; i++)  
            {   if ((people[i].custid == custid) && (people[i].custacctnum == custacctnum))
                {   match = true;
                System.out.println("Welcome "  +people[i].firstname+ " to JJ Dealership!");
                for(int p=0; p<cluster.length; p++)
                    System.out.println(+(p+1)+ ": " +cluster[p].year+"," +cluster[p].make+ "," +cluster[p].model);

                System.out.println(people[i].firstname+ ", what color car would you like?");
                    break;
                }
                }
                    if (!match)
                    {   System.out.println("Invalid ID");   
            }
            } catch (NumberFormatException e)
            {System.out.println ("Error! Please enter a number!");}
            } while (!(match));