Java 具有奇偶棋盘格的正负棋盘格

Java 具有奇偶棋盘格的正负棋盘格,java,Java,我如何检查输入的数字是负数还是正数,如果是正数,它将再次检查它是奇数还是偶数,如果是负数,将显示错误消息并再次询问用户是否要检查其他数字 import java.util.Scanner; public class OddEvenChecker{ public static void main(String[] args){ int integer=0; Scanner sc = new Scanner(System.in); whi

我如何检查输入的数字是负数还是正数,如果是正数,它将再次检查它是奇数还是偶数,如果是负数,将显示错误消息并再次询问用户是否要检查其他数字

import java.util.Scanner;

public class OddEvenChecker{
    public static void main(String[] args){
        int integer=0;

        Scanner sc = new Scanner(System.in);

        while (true) {
            if (integer>=0) {
                System.out.print("Enter a positive Integer: ");//gets the user input
                integer = sc.nextInt();
            }

            if (integer<0) { // checks if the entered number is a positive integer
                System.out.println("Error");
                System.out.print("Do you Want to Continue?  y for Yes, n for No: ");
                String choices = sc.nextLine();

            } else if(integer>0) {
                if (integer%2==0) {
                    System.out.println("Even");
                } else {
                    System.out.println("Odd");
                }
            }
        }
    }
}
import java.util.Scanner;
公共类检查程序{
公共静态void main(字符串[]args){
整数=0;
扫描仪sc=新的扫描仪(System.in);
while(true){
如果(整数>=0){
System.out.print(“输入正整数:”;//获取用户输入
整数=sc.nextInt();
}
if(整数0){
如果(整数%2==0){
System.out.println(“偶数”);
}否则{
系统输出打印项次(“奇数”);
}
}
}
}
}
公共类检查程序{

public static void main(String[] args){
    int integer=0;

    Scanner sc = new Scanner(System.in);

    while (true) {
        if (integer>=0) {
            System.out.print("Enter a positive Integer: ");
            integer = sc.nextInt();
            if(integer%2==0) {
                System.out.println("Even");
            }
            else {
                System.out.println("Odd");
            }
        }

        if (integer<0) {
            System.out.println("Error");
            System.out.print("Do you Want to Continue?  y for Yes, n for No: ");
            String choices = sc.nextLine();

        }   

   }


    }
publicstaticvoidmain(字符串[]args){
整数=0;
扫描仪sc=新的扫描仪(System.in);
while(true){
如果(整数>=0){
System.out.print(“输入正整数:”);
整数=sc.nextInt();
如果(整数%2==0){
System.out.println(“偶数”);
}
否则{
系统输出打印项次(“奇数”);
}
}
if(整数

您对这种方法有何看法?它是否满足要求?

提供的代码有什么问题?如果(整数>=0){
可以更改为
,而(整数)问题“是否要继续打印2次,并且我无法调整输入”如果用户输入的n不是“偶数”或“奇数”,我无法结束语句如果用户输入负数,则不应输出,否则不会输出。如果您愿意,请运行它。是的,但如果用户输入负数,则会提示您是否要继续,但我已经修复了它,非常感谢
public static void main(String[] args){
    int integer=0;

    Scanner sc = new Scanner(System.in);

    while (true) {
        if (integer>=0) {
            System.out.print("Enter a positive Integer: ");
            integer = sc.nextInt();
            if(integer%2==0) {
                System.out.println("Even");
            }
            else {
                System.out.println("Odd");
            }
        }

        if (integer<0) {
            System.out.println("Error");
            System.out.print("Do you Want to Continue?  y for Yes, n for No: ");
            String choices = sc.nextLine();

        }   

   }


    }
do{
choices = sc.nextLine();
if(choices.equals("n") || choices.equals("no"){
 return;
}
else if(choices.equals("y") || choices.equals("yes"){
 break;
} else {
System.out.println("invalid key"); 
 }
} while(!choices.equals("n") || (!choices.equals("no") || (!choices.equals("y") || (!choices.equals("yes"));
public class OddEvenChecker {
    public static void main(String[] args) {
        int integer = 0;
        String flag = "y";

        Scanner sc = new Scanner(System.in);
        Scanner intScanner = new Scanner(System.in);

        do {
            System.out.print("Enter a positive Integer: ");//gets the user input
            integer = intScanner.nextInt();

            if(integer>0) {
                if(integer%2==0) {
                    System.out.println("Even");
                } else {
                    System.out.println("Odd");
                }
            } else {
                System.out.println("This is negative number. Try again.");
            }

            System.out.print("Do you want to Continue?  y for Yes, n for No: ");
            flag = sc.nextLine();
        } while(!(flag.equals("n") || flag.equals("no")));
    }
}