如何保存数据库

时间:2017-08-03 07:20:55

标签: java mysql jdbc

我正在创建非常简单的ATM机。

开始时,系统会询问用户登录名和密码。 如果密码正确,则用户可以进行一些操作。

现在它的操作如下: 1)取钱 2)存钱 3)当前值

问题是,每当我选择1)或2)或3)时,程序会再次询问我有关登录名和密码的信息。但是我想在开始时只被问过一次。

所以我需要解决方案,它允许我保存用户登录名/密码,并在我想要进行某些操作时读取它。

这是我的代码的一部分。

public class LoginVerification {

public static void login() throws Exception {

    Scanner input = new Scanner(System.in);
    System.out.println("Please enter USER ID:");
    String userId = input.nextLine();
    System.out.println("Please enter PIN CODE:");
    String pass = input.nextLine();

    try {
        Connection con = ConnectionDB.getConnection();
        String sql = "SELECT COUNT(*) FROM Info WHERE ClientID = ? AND ClientPass = ?";

        PreparedStatement posted = con.prepareStatement(sql);
        posted.setString(1, userId);
        posted.setString(2, pass);
        ResultSet resultSet = posted.executeQuery();
        resultSet.next();
        int rowCount = resultSet.getInt(1);

        if (rowCount == 1) {
            System.out.println("LOGIN SUCCESSFUL");
            Helper.showMainMenu();

        } else {
            System.out.println("#LOGIN UNSUCCESSFUL, WRONG LOGIN OR PASSWORD");

            System.out.println("\nTry again? \n1) yes \n2) no [quit]");
            int choice = input.nextInt();
            if (choice == 1) {
                login();
            } else {
                System.exit(0);
            }
        }
        con.close();

    } catch (Exception e) {
        e.printStackTrace();
    }

}

}

问题如下。你可以看到该程序再次询问登录/通过,因为我不知道如何以不同的方式做到这一点。

public static void withdraw() throws Exception {

    Scanner input = new Scanner(System.in);
    System.out.println("please enter USER ID:");
    String userId = input.nextLine();
    System.out.println("please enter PIN CODE:");
    String pass = input.nextLine();

    System.out.println("how much you want to withdraw");
    int money = input.nextInt();

    try {
        Connection con = ConnectionDB.getConnection();
        String sql = "UPDATE BankDB.Info SET `Money`= `Money` - ? WHERE `ClientID`= ? AND `ClientPass` = ?";

        PreparedStatement posted = con.prepareStatement(sql);
        posted.setInt(1, money);
        posted.setString(2, userId);
        posted.setString(3, pass);

        if (posted.executeUpdate() > 0) {
            System.out.println("OPERATION SUCCESSFUL");
        } else {
            System.out.println("LOGIN UNSUCCESSFUL");

            System.out.println("\nTry again? \n1) yes \n2) no [quit]");
            int choice = input.nextInt();
            if (choice == 1) {
                withdraw();
            } else {
                System.exit(0);
            }
        }
        con.close();
    } catch (Exception e) {
        e.printStackTrace();
    }
}

1 个答案:

答案 0 :(得分:1)

你需要一个标志来告诉用户是否登录, 这可以在login()中设置,可以从任何页面检查/使用

修改login(),如下所示:

1-在LoginVerification class

中定义一个类范围变量
private static boolean loggedIn = false;

login()中的第一行,检查该标志

public static void login() throws Exception {

        if(loggedIn){
            //do nothing, user already logged in
            return;
        }
//...rest of code ...

当您发现用户登录是否正确时添加此内容:

if (rowCount == 1) {
    //add this line
    loggedIn = true;
} else {
    //add this line
    loggedIn = false;
}

现在在其他需要用户登录的地方,请使用

LoginVerification.login();

ex,替换此

public static void withdraw() throws Exception {

    Scanner input = new Scanner(System.in);
    System.out.println("please enter USER ID:");
    String userId = input.nextLine();
    System.out.println("please enter PIN CODE:");
    String pass = input.nextLine();

    System.out.println("how much you want to withdraw");// rest of code

用这个:

public static void withdraw() throws Exception {
    LoginVerification.login();
    System.out.println("how much you want to withdraw"); // rest of code
如果用户尚未登录,将要求用户登录。