Question

public class child {
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int AGE;
        System.out.println("Enter Child's number(s) of Days: ");
        AGE = scan.nextInt();

        if (AGE == 0 || AGE == 1){
            System.out.println("Classification: New Born");
        }
        else if (AGE >= 2 && AGE <= 10){
            System.out.println("Classification: Infant");
        }
        else if (AGE > 9 && AGE < 19){
            System.out.println("Classification: New Born");
        }
        else if (AGE > 17 && AGE < 37){
            System.out.println("Classification: TODDLER");
        }
        else if (AGE > 36 && AGE < 120 ) {
            System.out.println("Classification: KID");
        }
        else{
            System.out.println("Classification: Out of Range");
            System.out.println("Enter a number again:");
            AGE = scan.nextInt();
            if (AGE == 0 || AGE == 1){
                System.out.println("Classification: New Born");
            }
            else if (AGE >= 2 && AGE <= 10){
                System.out.println("Classification: Infant");
            }
            else if (AGE > 9 && AGE < 19){
                System.out.println("Classification: New Born");
            }
            else if (AGE > 17 && AGE < 37){
                System.out.println("Classification: TODDLER");
            }
            else if (AGE > 36 && AGE < 120 ) {
                System.out.println("Classification: KID");
            }           
            else{
                System.out.println("Classification: Out of Range");
                System.out.println("Enter a number again:");
                AGE = scan.nextInt();
            }   
        }   
    }
}

所以我根据他们的天数对孩子进行了分类。

如何简化此代码，因为如果用户输入的数字超出范围，程序将提示用户输入一个值，但如果用户再次输入一个超出范围的数字，将停止提示用户输入另一个号码，因为如果我在其他地方放置另一个条件构造，它将使我的代码更长，所以我的问题是如何使我的代码更短但是程序将继续提示用户，如果输入的数字是超出范围。

示例输出输入儿童的天数： 121 分类：超出范围再次输入一个数字： //程序会再次提示用户 -1 //但现在程序现在不会打印超出范围 / 如何让程序提示用户再次输入数字并制作我的课程更短 /

Answer 1

你应该熟悉一个while循环。

只要条件在每次迭代后保持为真，它就会循环。

Scanner scan = new Scanner(System.in);

while(true) {
    System.out.println("Enter Child's number(s) of Days: ");
    int AGE = scan.nextInt();

    if (AGE == 0 || AGE == 1) {
        System.out.println("Classification: New Born");
    } else if (AGE >= 2 && AGE <= 10) {
        System.out.println("Classification: Infant");
    } else if (AGE > 9 && AGE < 19) {
        System.out.println("Classification: New Born");
    } else if (AGE > 17 && AGE < 37) {
        System.out.println("Classification: TODDLER");
    } else if (AGE > 36 && AGE < 120) {
        System.out.println("Classification: KID");
    } else {
        System.out.println("Classification: Out of Range");
    }
}

另外，Java约定是用大写字母（child - ＆gt; Child）启动类名。

Answer 2

public class child {
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int AGE;
        while(true){
            System.out.println("Enter Child's number(s) of Days: ");
            try {
                AGE = scan.nextInt(); 
            } catch (Exception e) {
                System.out.println ("Please enter a number!");
            }            

            if (AGE == 0 || AGE == 1){
                System.out.println("Classification: New Born");
            }
            else if (AGE >= 2 && AGE <= 10){
                System.out.println("Classification: Infant");
            }
            else if (AGE > 9 && AGE < 19){
                System.out.println("Classification: New Born");
            }
            else if (AGE > 17 && AGE < 37){
                System.out.println("Classification: TODDLER");
            }
            else if (AGE > 36 && AGE < 120 ) {
                System.out.println("Classification: KID");
            }
            else{
                System.out.println("Classification: Out of Range");
            }
        }
    }
}

希望这有帮助。

如何让程序连续提示用户？

2 个答案: