Question

我是计算机编程的新手，我正在尝试创建一个程序，从用户的输入中读取最后三个字母，然后检查是否是解锁虚拟安全的正确代码。当用户输入“1”时，我似乎无法让程序关闭，这应该是退出的方式。在静态上下文中引用非静态变量“lock”也存在问题，有什么方法可以解决这个问题吗？感谢您的帮助。

/**
 * Gets the last three lettersof the user's input.
 * 
 * @author (Sean F.) 
 * @version (1.0)
 */
import java.util.Scanner;
public class ComboInput
{
public  static void main(String[] args)
{
    System.out.println("\f");
    Scanner in = new Scanner(System.in);
    System.out.print("Enter a letter");
    System.out.print("\n");
    String input = "";
    int comboLength = 0;
    input = in.nextLine();
    input = input.substring(0,1);
    comboLength ++;
    Lock q = new Lock();
    int lock = q.getLockStatus(input);
    String close = "1";
    while (comboLength < 3){
        System.out.println("\f");
        System.out.print("Your letters being used are:");
        System.out.print("\n");
        System.out.println(input);
        System.out.print("Enter another letter or type \"1\" to exit");
        System.out.print("\n");
        String newInput = in.nextLine();
        newInput = newInput.substring(0,1);
        input = input+newInput;
        comboLength++;
    }
    while (input.length()>=3){
        while((input.substring(2)).equals(close)){           
            System.out.println("\f");
            System.out.println("You gave up, lock is still locked");
            System.exit(0);
        }
        while(!((input.substring(2)).equals(close))){
            while (lock == 0){
                while (comboLength >= 3){
                    System.out.println("\f");
                    System.out.print("Your letters being used are:");
                    System.out.print("\n");
                    System.out.println(input.substring(comboLength-3));
                    System.out.print("Enter another letter or type \"1\" to exit");
                    System.out.print("\n");
                    String newInput = in.nextLine();
                    newInput = newInput.substring(0,1);
                    input = input+newInput;
                    comboLength++;
                    int lock2 = q.getLockStatus(input);
                    lock = lock2;
                }  
            }
            while (lock == 1){
                System.out.println("\f");
                System.out.println("Your Combination is Correct. Unlocked");
                System.exit(0);
            }
        }
    }
}

}

这是该计划的第二部分。这是一个名为Lock的类

/**
 * Gives the current status of the lock
 */
public class Lock
{
private String code;
/**
 * Constructs the correct code for the lock
 */
public Lock()
{
    code = "smf";
}

/**
 * 
 * 
 * @return     An integer to describe whether lock is open or not
 */
public int getLockStatus(String c)
{
        if (c.equalsIgnoreCase(code)){
            return 1;
        }
        else{
            return 0;
        }       
}

Answer 1

如果用户输入1，我肯定看不到任何要退出的代码，即使提示说明了这一点。您可以通过在newInput = newInput.substring(0,1);之后添加此代码来解决此问题：

if (newInput.equals(close))
    System.exit(0);

顺便说一句，该代码中包含大约20个不同的错误。我将为您省去麻烦并为您编写代码：

import java.util.Scanner;

public class ComboInput {
    Scanner in = new Scanner(System.in);
    System.out.print("Please enter the combination, or 1 to exit: ");
    String combo = in.readLine();
    if (combo.equals("1"))
        System.exit(0);
    if (combo.equals("smf")) 
        System.out.println("That is the correct combination. The safe is open.");
    else 
        System.out.println("Incorrect combination.");
}

研究这段代码，找出它的工作原理，并将其作为编写简单代码的一课。

如果您坚持使用Lock类，则只需更改一行。看看你是否能搞清楚。

锁定程序非静态变量问题，以及关闭问题

1 个答案: