Question

我有点球大战的Java程序，我想将b1，b2，b3和s1限制为1-9之间的任何数字。如果输入不等于1-9，请再试一次。

 for (int i=0;i<=5;i++)

        {

            int b1, b2, b3, s1;
            int j=i+1;
            System.out.println("Enter the Numbers which you will block.... Please remember that it should be from 1-9 and nothingelse, 1 2 3 4 5 6 7 8 9");


            b1=Integer.parseInt(br.readLine()); // how to restrict this to numbers between 1 to 9
            b2=Integer.parseInt(br.readLine()); //this also
            b3=Integer.parseInt(br.readLine()); //this also

            System.out.flush();
            System.out.println("Enter The Number where you will score");
            s1=Integer.parseInt(br.readLine()); //this also

Answer 1

伪代码：

if ((input > 9) || (input < 1)) {
    tryAgain()
}

我认为这就是你所需要的（你必须检查每个变量）。

Answer 2

试试这种动态方式，

public class testJava {
   public static void main(String[] args) {
       testJava j = new testJava();
       System.out.println(j.isBetween(1,1,9));

   }
    public boolean isNumeric(int value,int startValue,int endValue) {
        if(value >= startValue || value <= endValue)
        {
            return true;
        }else{
           return false;
        }
    }
}

Answer 3

只需检查输入的输入。

If (number >= 1 && number <= 9)
{
 progressFurther();
}
else
{
 enterAgain();
 checkAgain();
}

Answer 4

只需使用简单的条件，例如：

if (x >= 1 && x <= 9) {
    // handle the correct case
} else {
    // handle the case where the input is not in the range 1-9
}

Answer 5

while(b1 > 9 || b1 < 1) {
    System.out.println("Wrong number. Try again."); 
    b1=Integer.parseInt(br.readLine());
}

如何将输入整数从1限制为9

5 个答案: