Question

我正在尝试创建一个检查用户密码的程序。我希望程序在用户正确的情况下结束，但如果不是，我希望它只询问4次。

问题：即使您确实正确获取了密码，程序也会继续询问猜测密码。如果你弄错了，它会错误地提问。我该如何解决这个问题？

import java.util.Scanner;
public class HowToAdd {
    public static void main (String [] args){
        Scanner input = new Scanner (System.in);
        int trys = 0;
        String password=null;
        do{
            System.out.println("Guess the password");
            password = input.next();

            if(password.equalsIgnoreCase("Password")){
                System.out.println("Great job");
            }else{
                System.out.println("Try again");
                input.next();
                trys++;
            }
        }while(trys<2);
        System.out.println("Try again later!");

        input.close();
    }
}

Answer 1

你只需要添加一个休息：

if(password.equalsIgnoreCase("Password")){
 System.out.println("Great job");
 break;
}

Answer 2

我不仅添加了一个中断来修复输入正确密码时不离开循环的问题，而且还添加了一些其他内容来帮助您查看以下内容：

 public static void main (String [] args){
    Scanner input = new Scanner (System.in);
    int trys = 0;
    String password=null;
    System.out.println("Guess the password:");
    while(trys<2)
    {
        password = input.next();

        if(password.equalsIgnoreCase("Password")){
            System.out.println("Great job");
            break;
        }else{
            trys++;
            if(trys != 2)
            {
                System.out.println("Try again:");
            }
        }
    }
    if(trys == 2)
    {
        System.out.println("Try again later!");
    }

    input.close();
}

尝试这一点，如果使用break;语句正确，它将突破循环。此外，它只会在第一次尝试时显示猜测密码，然后再尝试。如果他们猜对了，也不会再说再试，因为它会检查他们是否猜错了两次。

Answer 3

您可以使用break关键字进行修复，例如：

        if(password.equalsIgnoreCase("Password")){
            System.out.println("Great job");
            break;
        }

此处解释了break：

Branching Statements

Answer 4

您可以在不使用break语句的情况下完成所需的操作。试着看看我的代码流程。

    Scanner scn = new Scanner(System.in);
    int tries=0;
    String pw = "";

    System.out.println("Guess the password");
    pw = scn.nextLine();

    while(!pw.equalsIgnoreCase("Password") && tries < 3){
        tries++;
        System.out.println("Try again");
        pw = scn.nextLine();            
    }

    if(tries >= 3)
        System.out.println("Try again later!");
    else
        System.out.println("Great job!");

TEST RUN 1：

Guess the password
1st try
Try again
2nd try
Try again
3rd try
Try again
last try
Try again later!

TEST RUN 2：

Guess the password
password
Great job!

Answer 5

以下是代码的另一种变体：

    public static void main(String[] args) {
    Scanner input = new Scanner(System.in);
    int trys = 0;
    String password = null;
    // create variable for a number of tries
    int numberOfCheks = 4;
    do {
        System.out.println("Guess the password");
        password = input.nextLine();

        if (password.equalsIgnoreCase("Password")) {
            System.out.println("Great job");
            break;
        } else if (++trys == numberOfCheks) {
           //Break if you've used all your tries 
            System.out.println("Try again later!");
            break;
        } else {
            System.out.println("Try again");
            // input.next(); - Useless. You don't handle it. Removed
        }
    //There is already tries number check. So 'true' can be used as the condition.
    } while (true); 
    input.close();
}

密码检查程序java程序

5 个答案: