Question

请参阅下面的代码我试图在成功登录后调用Jmenu类

登录：

public class Login {

    Connection con;
    Statement st;
    ResultSet rs;

    JFrame f = new JFrame ("User Login");
    JLabel l = new JLabel ("UserName:");
    JLabel l1 = new JLabel ("Password:");
    JTextField t = new JTextField (10);
    JTextField t1 = new JTextField (10);
    JButton b = new JButton ("Login");



    public Login ()
    {
        connect ();
        frame ();
    }

    public void connect ()

    {
        try
        {
        String driver = "sun.jdbc.odbc.JdbcOdbcDriver";
        Class.forName(driver);

        String db = "jdbc:odbc:Joy_DB";
        con = DriverManager.getConnection(db);
        st = con.createStatement ();
        }
        catch (Exception ex)
        {


        }
    }

    public void frame ()
    {

        f.setSize (600,400);
        f.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
        f.setVisible (true);

        JPanel p = new JPanel ();
        p.add (l);
        p.add (t);
        p.add (l1);
        p.add (l);
        p.add (t1);
        p.add (b);

        f.add (p);
        b.addActionListener(new ActionListener(){

            public void actionPerformed (ActionEvent e)

            {
                try
                {
                String user = t.getText (). trim ();
                String pass = t1.getText (). trim ();

                String sql = "select User,Password from Table2 where User = '"+user+"' and Password = '"+pass+"'";
                rs = st.executeQuery(sql);

                int count = 0;

                while (rs.next())
                {
                    count = count +1;

                }

                if (count == 1 )
                {
                    JOptionPane.showMessageDialog(null,"User Found");
                    //JMenuDemo M = new JMenuDemo ();
                }

                else if (count > 1)
                {
                    JOptionPane.showMessageDialog(null, "Duplicate User !");
                }

                else
                {
                    JOptionPane.showMessageDialog (null,"User does not exist");
                }

                }
                catch (Exception ex)
                {

                }
            }
    });


    }
    /**
    * @param args the command line arguments
    */
    public static void main(String[] args) {

        new Login ();
        //JMenuDemo M = new JMenuDemo ();
        // TODO code application logic here
    }
}

使用上面的代码

成功登录后如何调用J菜单框

请帮忙我将发送另一个类是Jmenu ia一段时间

Answer 1

使用收集用户详细信息所需的所有逻辑定义LoginPanel
创建另一个包含应用程序组件和逻辑的面板。
使用JDialog显示登录面板。在对话框关闭之前，它将阻止代码的执行
根据LoginPane的状态，您（可能）退出应用程序（登录失败）或继续运行应用程序。
将应用程序面板添加到JFrame并使其可见

有关详细信息，请参阅How to make dialogs。

您可能还想使用PreparedStatement来访问数据库中的信息

如何在java swing中成功登录后调用Jmenu屏幕

1 个答案: