Question

我正在尝试在此程序中实现登录功能。我终于弄明白了如何做一个基本的，但遗憾的是我不知道如何结束它，例如如果用户最终达到3的限制它应该结束，但我仍然继续，我不知道我应该放在哪里和什么代码，以便它结束而不是继续主程序。

import java.io。*;

public class Password {

public static void main(String args[]) throws IOException{

    String name, un, pw;
    String Username = "passwordtest";
    String Password = "test123";
    int stud;
    double math, science, english, filipino, social, ave, sum, fingrade;

    BufferedReader inpt = new BufferedReader (new InputStreamReader(System.in));

    for(int trial=1; trial<=3; trial++){
    System.out.print("Username: ");
    un = inpt.readLine();

    System.out.print("Password: ");
    pw = inpt.readLine();

    System.out.println("");

    if (un.equals(Username) && pw.equals(Password)){
        System.out.println("You have successfully logged in!");
        trial=trial+2;
        continue;
    }else{
        System.out.println("Sorry, Incorrect Username/Password");
        System.out.println("Please Try Again");
        System.out.println("");
        }
    }
    System.out.println("");

    System.out.println("Welcome to ITMinions' Grading System!");
    System.out.println("How many students' grades would you like to record?");
    System.out.print("Answer: ");
    stud=Integer.parseInt(inpt.readLine());

    System.out.println("");

for (int ctr=1; ctr<=stud; ctr++){
    System.out.print("Name of the student: ");
    name = inpt.readLine();

    System.out.println("");
    System.out.println("Input the following grades");

    System.out.print("Math: ");
    math = Double.parseDouble(inpt.readLine());

    if(math<65 || math>100){
        System.out.println("");
        System.out.println("Wrong input, try again.");
        System.out.println("");
        ctr=ctr-1;
        continue;
    }

    System.out.print("Science: ");
    science = Double.parseDouble(inpt.readLine());

    if(science<65 || science>100){
        System.out.println("");
        System.out.println("Wrong input, try again.");
        System.out.println("");
        ctr=ctr-1;
        continue;
    }

    System.out.print("English: ");
    english = Double.parseDouble(inpt.readLine());

    if(english<65 || english>100){
        System.out.println("");
        System.out.println("Wrong input, try again.");
        System.out.println("");
        ctr=ctr-1;
        continue;
    }

    System.out.print("Filipino: ");
    filipino = Double.parseDouble(inpt.readLine());

    if(filipino<65 || filipino>100){
        System.out.println("");
        System.out.println("Wrong input, try again.");
        System.out.println("");
        ctr=ctr-1;
        continue;
    }

    System.out.print("History: ");
    social = Double.parseDouble(inpt.readLine());

    if(social<65 || social>100){
        System.out.println("");
        System.out.println("Wrong input, try again.");
        System.out.println("");
        ctr=ctr-1;
        continue;
    }

    sum=math+science+english+filipino+social;
    ave=sum/5;

    System.out.println("");
    System.out.println("The average of " + name + " is: " + ave);
    System.out.println("");

 }

}

}

请帮忙！是的，这与学校工作有关:) 谢谢！

Answer 1

我会重写处理成功登录的循环部分，如下所示：

    if (un.equals(Username) && pw.equals(Password)){
        System.out.println("You have successfully logged in!");
        break;
    }

请注意使用break关键字来摆脱循环。

当用户使用所有登录尝试时，您可以使用System.exit(0);退出。

Answer 2

您必须使用另一个变量，例如：boolean isLoggedIn，并设置成功登录，然后中断而不是继续，如下所示：

if (un.equals(Username) && pw.equals(Password)){
    System.out.println("You have successfully logged in!");
    isLoggedIn = true;
    break;
}else{
    System.out.println("Sorry, Incorrect Username/Password");
    System.out.println("Please Try Again");
    System.out.println("");
}

然后在for循环之外，检查是否（isLoggedIn）并相应地执行操作。

Answer 3

修改你的fi声明

if (un.equals(Username) && pw.equals(Password)){
        System.out.println("You have successfully logged in!");
        break;;
    }else{
        System.out.println("Sorry, Incorrect Username/Password");
        System.out.println("Please Try Again");
        System.out.println("");
        }
    }


if(trail==4)
{
   //Write your locking logic here
}

在代码中对代码进行硬编码并不是一种好习惯。尝试使用属性文件以简化或如果您有时间使用jdbc

要避免使用Null指针异常

Username.equals(un)

Answer 4

另外，请确保遵循适当的Java编码标准，例如用于变量命名的camel-case，以及用于常量的所有大写字母。由于用户名和密码是硬编码的，因此它们是非常常见的。所以，改变

String Username = "passwordtest";
String Password = "test123";

到

final String USERNAME = "passwordtest";
final String PASSWORD = "test123";

如果您可以从属性文件加载这些常量也会更好，因为当密码更改时，您无需修改代码，只需编辑属性文件。

Answer 5

澄清以前的答案：

booelan isLoggedIn = false;
for ( int trials = 3; trials > 0; trials-- )
{
  <ask uname, password> // Java convention: don't capitalise variable names
  if ( isLoggedIn = <uname/password are OK> {
    System.out.println ( "Success" );
    break;
  }

  System.out.printf ( "Bad uname/pass, %d attempts remaining\n", trials );
}

if ( !isLoggedIn ) {
  System.out.println ( "User couldn't give valid credentials, quitting after three attempts, due to security reasons" );
  Thread.sleep ( 3000 ) // try to fight brute-force attackers 
  System.exit ( 1 ); // Not zero, it's not a regular end 
}

// Go ahead with your application

Java基本登录功能

5 个答案: