Question

我有这个评分程序，我正在尝试使用键盘输入退出程序的很多东西，在这种情况下是字母“E”。我创建了另一种方法，但我不能让它与main方法一起工作，而且我是如此新，以至于我被困住了。我尝试使用布尔值，并使用字符串转换为int，但我得到的只是错误。我的代码在下面，提前谢谢你，我知道这可能是一个愚蠢的问题。

import java.util.Scanner;
public class grading {

    public static void main (String args[]){
        boolean run = true;//this code together with bottom keep loop running
        while(run){ //the code inside this brackets will continue the loop

            int yourScore;
            Scanner input = new Scanner(System.in);
            System.out.println("Enter your Score: ");
            yourScore = input.nextInt();

            boolean kill;
            System.out.println("Press E to Exit");
            Scanner input0 = new Scanner(System.in);

            if (yourScore < 60 && yourScore >=0){
                System.out.println("You had an F, Sorry!");
            }
            else if (yourScore >=60 && yourScore < 70){
                System.out.println("You had an D, Study more!");
            }
            else if (yourScore >=70 && yourScore < 80){
                System.out.println("you had an C, you can do Better!");
            }
            else if (yourScore >= 80 && yourScore < 90){
                System.out.println("You had an B, very Well done!");
            }
            else if ( yourScore >=90 && yourScore <= 100){
                System.out.println("you had an A, you are Great!");
            }
            else if (kill = "e" != null){
                System.out.println("Bye!");
                Runtime.getRuntime().exit(0);
            }
            else
                System.out.println("Incorrect Input!");
        }
    }
}

Answer 1

您的代码中有几处错误。首先，您允许用户输入数字和字母，但将输入分配给您的分数是int。如果用户输入一个字母，那么您试图将该字母指定为整数，这会抛出InputMismatchException。

其次，您正在使用两个扫描仪，但从不使用第二个扫描仪（即输入0）。实际上，您在此程序中不需要两个Scanner。只有一台扫描仪可以完成这项工作。

第三，在最后一个if语句中，你要分配kill =＆＃34; e＆＃34; ！= null。我很确定这不是你打算做的。您应该首先熟悉java的工作原理。在java中，执行从右到左进行。在这里表达＆＃34; e＆＃34;首先评估！= null。因为＆＃34; e＆＃34;是一个字符串，它永远不会等于null。因此，表达＆＃34; e＆＃34; ！= null将评估为true。然后将此真值分配给kill变量。因此，无论如何，kill变量总是如此。这不是你想要的。以下是可以按预期工作的代码。

public class grading {
public static void main (String args[]){
    boolean run = true;//this code together with bottom keep loop running
    while(run){ //the code inside this brackets will continue the loop

        int yourScore;
        Scanner input = new Scanner(System.in);
        System.out.println("Enter your Score or E to exit: ");
        //checking if the input is integer or not
        if(input.hasNextInt()){ 
            yourScore = input.nextInt();

            if (yourScore < 60 && yourScore >=0){
                System.out.println("You had an F, Sorry!");
            }
            else if (yourScore >=60 && yourScore < 70){
                System.out.println("You had an D, Study more!");
            }
            else if (yourScore >=70 && yourScore < 80){
                System.out.println("you had an C, you can do Better!");
            }
            else if (yourScore >= 80 && yourScore < 90){
                System.out.println("You had an B, very Well done!");
            }
            else if ( yourScore >=90 && yourScore <= 100){
                System.out.println("you had an A, you are Great!");
            }
            else
                System.out.println("Incorrect Input!");
        }
        // checking if the input is string
        else if(input.hasNext()){
            //assigning input to String varible
            String userInput = input.next();
            //comparing the input value with letter e ignoring the case
            if(userInput.equalsIgnoreCase("e")){
                run = false;
            }
            else{
                System.out.println("Invalid key. Press e to exit");
            }
        }

    }
}
}

.java中的结束程序使用字母“e”将字符串转换为int以退出程序

1 个答案: