Question

我正在为一篇课程写一个假期推荐系统。其GUI使用CardLayout。在主类中，创建一个用户对象，其默认名称和访问级别在其构造函数中定义。此对象从main传递到UserCard面板，该面板将其传递给Login并登录。如果用户成功登录，则cardpanel从Login转换为登录，并且应该通过调用user.getUsername（）来显示登录用户的用户名;方法

因此我的问题。由于卡布局的工作方式，具有用户名显示的面板已经在UserCards的构造函数中创建，其默认值从那时起首先创建了用户对象。我需要找到一种方法来强制此面板在cardlayout对象上调用show方法后重新绘制。以下是有问题的3个类的代码。（我已将代码粘贴限制在相关方法中。）

//the usercards panel

        public UserCards(User u)
    {
        CardLayout cl = new CardLayout();
        this.setLayout(cl);

        UserOptionsPanel options_card = new UserOptionsPanel(cl, this);
        RegisterPanel register_card = new RegisterPanel(cl, this);
        LoggedInPanel loggedin_card = new LoggedInPanel(cl, this, u);
        LoginPanel login_card = new LoginPanel(cl, this, u, loggedin_card);

        this.add(options_card, options);
        this.add(login_card, login);
        this.add(register_card, register);
        this.add(loggedin_card, loggedin);
    }

//the Loggin action listener user is passed in as a reference to the user object created in //main. the createUser(); method is a badly named method that simply calls setter methods on //the user object's fields

    @Override
    public void actionPerformed(ActionEvent e) 
    {
        String[] vals = packData();
        try
        {
            DBConnection d = new DBConnection();
            Connection conn = d.getConnection();
            Validation v = new Validation(vals, user);
            v.getActual(conn);
            if(v.validate())
            {
                user = v.createUser();
                System.out.println(user.getUserName());
                l.revalidate();
                cl.show(pane, "loggedin");
            }
            else
            {
                lbl_statusmsg.setText("Password Incorrect");
                lbl_statusmsg.repaint();
            }

        }
        catch(ClassNotFoundException | SQLException ex)
        {
            ex.printStackTrace();
        }
    }

//the loggedin constructor

public class LoggedInPanel extends JPanel
{
    private User user;
    private JLabel lbl_details;
    public LoggedInPanel(CardLayout cl, Container pane, User u)
    {
        super();
        user = u;
        this.setLayout(new BoxLayout(this, BoxLayout.Y_AXIS));

        lbl_details = new JLabel();
        lbl_details.setText("Welcome "+user.getUserName());

        this.add(lbl_details);
    }    
}

道歉，如果我没有过分清楚我不会被要求帮助：）

Answer 1

你的意思是什么？

CardLayout cl = new CardLayout() {
  @Override
  public void show(java.awt.Container parent, String name) {
    super.show(parent, name);
    // your code here

  }
};

在调用CardLayout的show方法后强制JPanel重绘

1 个答案: