Java For 循环数据验证

Question

我目前正在大学里做一些 Java 练习，我已经在这个练习上停留了大约 5 个小时了!我正在练习 For 循环并让循环询问 5 次 1 到 3 的数字。测试时，如果我输入无效选择，它会继续并将无效选择包含为零，当输入了无效输入，但它仍然会继续，直到循环结束，我知道有一种方法可以返回到选择的开始，但我无法弄清楚。 我到处寻找解决方案，但找不到!我知道这不会太多，而且我已经几天没有回到大学了，所以我不能问讲师，我真的很想继续学习下一章。

这是我的代码(我知道它可能有点乱!!)，谢谢，Rob

   import java.util.Scanner;

/* this is s a survey of how 5 people sweeten thier coffee */

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

        int person, preference, nothing, sugar, sweetner;

        String pluralone = "People dont";
        String pluraltwo = "People use";
        String pluralthree = "People use";

        person = 0;
        preference = 0;
        nothing = 0;
        sugar = 0;
        sweetner = 0;

        for (person = 1; person <= 5; person++)

        {
            System.out.println("How do you sweeten your coffee");
            System.out.println("1. I Don't");
            System.out.println("2. With Sweetner");
            System.out.println("3. With Sugar");

            preference = input.nextInt();

            if (preference != 1 && preference != 2 && preference != 3)
                System.out.println("Sorry that is not a valid option");

            else if (preference == 1)
                nothing++;

            else if (preference == 1)
                sweetner++;

            else
                sugar++;
        }

        System.out.println("Survey Report");

        System.out.println("#############");

        if (nothing < 2)

        {
            pluralone = "person doesnt";
        }

        System.out.println(nothing + "  " + " " + pluralone + " sweeten thier coffee");

        if (sweetner < 2)

        {
            pluraltwo = "person uses";
        }

        System.out.println(sweetner + "  " + pluraltwo + " " + "sweetner to sweeten thier coffee");

        if (sugar < 2)

        {
            pluralthree = "person uses";
        }

        System.out.println(sugar + "  " + pluralthree + " " + "sugar to sweeten thier coffee ");

    }
}

Answer 1

只需在一个 while 循环中要求用户选择，这样它就不会继续，直到输入了一个有效的选项，比如:

preference = input.nextInt();
while (preference != 1 && preference != 2 && preference != 3) {
    System.out.println("Sorry that is not a valid option");
    preference = input.nextInt();
}

或者，您可以减少 if 语句中的 person 以引起 for 循环的另一次迭代，但这有点 hacky:

if (preference != 1 && preference != 2 && preference != 3) {
    System.out.println("Sorry that is not a valid option");
    person--;
}