你不必扫描键盘上的数字

根据要求，您不必扫描键盘上的数字；相反，你必须从一个自然数序列中扫描它们。因此 Scanner 对象需要用自然数序列示例化，而不是 System.in .

import java.util.Scanner;

class Main {
    public static void main(String[] args) {
        String nums = "4 10 45 60 63 75 78 90 92 0 15 18";
        Scanner scanner = new Scanner(nums);
        while (scanner.hasNextInt()) {
            int n = scanner.nextInt();
            if (n == 0) {
                break;
            }
            System.out.println(n + " => " + (n % 2 == 0 ? "Even" : "Odd"));
        }
    }
}

输出：

4 => Even
10 => Even
45 => Odd
60 => Even
63 => Odd
75 => Odd
78 => Even
90 => Even
92 => Even

更新（基于op的澄清）

您在评论中提到，您的要求确实是从键盘扫描数字。除此之外，您还提到要验证这些数字。
为此，可以使用无限循环（例如。 while(true) { } )当输入为 0 . 如果数字无效，只需显示一条错误消息，而不进行处理。
演示：

import java.util.InputMismatchException;
import java.util.Scanner;

class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n;
        while (true) {
            try {
                n = scanner.nextInt();
                if (n == 0) {
                    break;
                } else if (n < 0) {
                    System.out.println("It's not a natural number. Please try again.");
                } else {
                    System.out.println(n + " => " + (n % 2 == 0 ? "Even" : "Odd"));
                }
            } catch (InputMismatchException e) {
                System.out.println("It's not an integer. Please try again.");
            }
        }
    }
}

示例运行：

10
10 => Even
-2
It's not a natural number. Please try again.
5
5 => Odd
15
15 => Odd
16
16 => Even
-30
It's not a natural number. Please try again.
0

public static void main(String[] args) {

    Scanner scanner = new Scanner(System.in);

    int number = 0;

    System.out.println("Start putting numbers");

    while (scanner.hasNext()) {
        number = scanner.nextInt();
            if (number % 2 != 0) {
                System.out.println("Odd");
                continue;
            } else if (number % 2 == 0){
                System.out.println("even");
            }
            if(number == 0)
                break;
            }
        }
    }
}

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

    while (true) {
        int num = scan.nextInt();

        if (num == 0)
            break;

        System.out.println(num % 2 == 0 ? "even" : "odd");
    }
}

我认为您的代码中有一个输入错误：）请重构行：

System.out.println("Odd");

到

System.out.println("odd");

此外，您可能根本不调用循环控件（break，continue）：

public static void main(String... args) {
    Scanner scanner = new Scanner(System.in);
    int number = 0;

    while ((number = scanner.nextInt()) != 0) {
        if (number % 2 != 0) {
            System.out.println("odd");
        } else {
            System.out.println("even");
        }

    }
}

当数字是偶数时，你不应该打破这个条件。应该有另一个条件来检查 number = 0 ```
public static void main(String[] args) {
// put your code here
Scanner scanner = new Scanner(System.in);
int number = 0;

    while (scanner.hasNext()) {
        number = scanner.nextInt();
        if(number == 0) {
             break;
        }
        else if (number % 2 == 0) {
            System.out.println("even");
        } else {
            System.out.println("odd");
        }
   }

}