Question

我有一个Java分配来制作如下评分等级。

 import java.util.Scanner;
public class gradeSelection {
   public static void main(String[] args){
   Scanner input = new Scanner(System.in);
   System.out.print("Enter Score: ");
   double score = input.nextDouble();
   if (score >= 90) {
        System.out.println("Your score is " + score + " which is an A");
        if ((score < 90) && (score >= 80)){
            System.out.println("Your score is " + score + " which is an B");
            if ((score < 80) && (score >= 70)){
                System.out.println("Your score is " + score + " which is an C");
                if ((score < 70) && (score >= 60)){
                    System.out.println("Your score is " + score + " which is an D");
                    if (score < 60){
                        System.out.println("Your score is " + score + " which is an E");
                    }//endif
                }//endif    
            }//endif
        }//endif
    }//endif
}
}

我可以让它在Eclipse中工作但它会在我输入数字后终止。我究竟做错了什么？

Answer 1

您必须使用if和else if，因为如果第一个条件不匹配，它将退出并且不会达到嵌套（其他）条件：

尝试将其更改为：

if (score >= 90) {
    System.out.println("Your score is " + score + " which is an A");
} else if (score >= 80){
    System.out.println("Your score is " + score + " which is an B");
}//and so on

Answer 2

很简单，如果输入不大于90，则不打印任何内容。使用else if而不是这种方法。

Java中的评分量表

2 个答案: