Question

我进行了程序测试，看看用户输入是正数还是整数。 if语句测试是否为整数，else if测试是否为负数。如果是负数或小数，则要求用户输入正整数。问题出在else语句，它正在等待用户输入，我希望它使用System.out.print("Enter the test number: ");中的值，如果它传递if和else if测试。

我尝试将System.out.println("Please enter an integer!");之后的用户输入分配给一个int变量，但是如果用户输入一个double，我会得到一个错误，所以我认为这种方式不起作用。任何关于如何使程序工作的见解表示赞赏，谢谢！

import java.util.Scanner;  

public class FibonacciNumbersTester  
{  
    public static void main(String[]args)  
    {   //Variables  
        Scanner userDed = new Scanner(System.in);  
        String userChoice = "Y";  
        while(userChoice.equalsIgnoreCase("Y"))  
        {  
            Scanner userNum = new Scanner(System.in);  
            System.out.print("Enter the test number: ");  
            if(!userNum.hasNextInt() )  
            {  
                System.out.println("Please enter an integer!");  
            }  
            else if(userNum.nextInt() < 0 )  
            {     
                System.out.println("Please enter a postive integer!");  
            }  
            else  
            {  
                int NumTo = userNum.nextInt();  
                System.out.println(NumTo);  
            }  

            System.out.print("Would you like to continue? (Y/N)");  
            userChoice = userDed.next();                
        }  
    }  
}

感谢。

Answer 1

您应该拨打nextInt一次，保存结果并将其用于比较。

试试这个：

public static void main(String[] args) { // Variables
        Scanner userDed = new Scanner(System.in);
        String userChoice = "Y";
        while (userChoice.equalsIgnoreCase("Y")) {
            Scanner userNum = new Scanner(System.in);
            System.out.print("Enter the test number: ");
            if (!userNum.hasNextInt()) {
                System.out.println("Please enter an integer!");
            } else {
                int NumTo = userNum.nextInt();
                if (NumTo < 0)
                    System.out.println("Please enter a postive integer!");
                else
                    System.out.println(NumTo);

            }
            System.out.print("Would you like to continue? (Y/N)");
            userChoice = userDed.next();

        }
    }

Answer 2

Pattern positiveInt = Pattern.compile("^[1-9]\d*$"); // for positive integer
if(!userNum.hasNext(positiveInt)) {
    System.out.println("Please enter an positive integer (greater than 0) !");  
}
else {
    int NumTo = userNum.nextInt(positiveInt);
    System.out.println(NumTo);  
}

Java if-else比较

2 个答案: