扫描器+开关使用

Question

我是Java的新手。

我目前正在做一个辅助项目，制作一个基于文本的游戏。我意识到使用开关语句对这种类型的游戏非常有用。

所以这基本上就是它的工作原理。

我问用户，你想做什么？

1. 吃
2. 步行
3. 等。

那么，与“再次询问用户的默认语句”一起构建switch-statement和Scanner的最佳方法是什么？

我一直在这样做(我的代码在这里)，但它似乎有很多潜在的问题。

你们能给我一些关于如何用switch-statement制作最佳Scanner的提示吗？

先谢谢你。

public static void ask() {
 Scanner sc = new Scanner(System.in);
    System.out.println("What do you want to do?");
    while (!sc.hasNextInt()) {
        sc.next();
    }
        select = sc.nextInt();
        switch (select) {

        case 1:
            eat();
            break;
        case 2:
            walk();
            break;
        case 3:
            sleep();
            break;

        default:
            System.out.println("choose from 1 to 3");
           ask();  //would you re call itself again here? or is there any otherway to do without recalling itself?

        }

Answer 1

您的代码看起来不错，但我会重构它。还添加while循环以再次询问：

public static void ask() {

 Scanner sc = new Scanner(System.in);
 boolean isWrongAnswer;
 do {
    isWrongAnswer = false;
    System.out.println("What do you want to do?");
    switch (sc.nextInt()) {
        case 1:
            eat();
            break;
        case 2:
            walk();
            break;
        case 3:
            sleep();
            break;
        default:
           System.out.println("choose from 1 to 3");
           isWrongAnswer = true;
        }
 } while (isWrongAnswer);
}

Answer 2

最好先检查扫描仪是否有什么东西，然后检查它是否为int值。为此，可以使用sc.hasNext() & sc.hasNextInt()，如下所示，

 public static void ask() {
            Scanner sc = new Scanner(System.in);
            System.out.println("What do you want to do?");
            while (sc.hasNext()) {
                int select = 0;
                if (sc.hasNextInt()) {
                    select = sc.nextInt();
                }
                switch (select) {

                case 1:
                    System.out.println("call eat()");
                    break;
                case 2:

                    System.out.println("call walk()");
                    break;
                case 3:
                    System.out.println("call sleep()");
                    break;

                default:
                    System.out.println("choose from 1 to 3");
                    ask(); 

                }
            }

Answer 3

我建议你在这种情况下使用字符串。以提高代码的可读性，并避免错误。(http://www.oracle.com/technetwork/java/codeconventions-150003.pdf)

private static final String EAT = "1";
private static final String WALK = "2";
private static final String SLEEP = "3";

public void ask() {
    Scanner sc = new Scanner(System.in);
    System.out.println("What do you want to do?");
    switch (sc.next()) {
        case EAT:
            eat();
            break;
        case WALK:
            walk();
            break;
        case SLEEP:
            sleep();
            break;
        default:
            System.out.println("choose from 1 to 3");
            ask();  // as described above via "do while", but there is nothing wrong with recursion. The garbage collector works.
    }
}