Java Basic登录特性

Question

我试图在这个程序中实现一个登录功能。我终于想出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，如果用户最终达到了3的极限，它应该结束，但我的仍然继续，我不知道我应该在哪里和什么代码，以便它结束比继续主程序。

进口java.io.*；

公共类密码{

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("");

 }

}

}

请帮帮我！是的，这与学校的工作有关:)谢谢！

Answer 1

我将重写循环中处理后续完整登录的部分，如下所示：

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

注意，使用break关键字可以从循环中分离出来。

当用户使用了所有登录尝试时，您可以使用System.exit(0);退出。

Answer 2

例如，您必须使用另一个变量:布尔isLoggedIn，如果成功登录，则设置该变量，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if(isLoggedIn)并执行相应的操作。

Answer 3

修改fi语句

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
}

在代码中硬编码并不是很好的实践。为了简单起见，尝试使用属性文件，或者如果有时间使用jdbc。

若要避免空指针异常，请使用

Username.equals(un)