Question

抱歉我的英文不好......我为我的Java GUI程序创建了一个登录界面。我停留在仪表板框架上...我不知道如果我点击登录按钮，它将被移动到一个新的框架...如果我进行此注册并仅在1个源java中登录就可以了文件？

全部，谢谢

这是我的代码

import java.awt.event.KeyListener;

import java.awt.event.MouseEvent;
import java.awt.event.MouseListener;
import java.awt.BorderLayout;
import java.awt.EventQueue;
import java.awt.GridLayout;
import java.awt.event.ActionEvent;
import java.awt.event.ActionListener;
import java.awt.event.KeyEvent;

import javax.swing.ImageIcon;
import javax.swing.JButton;
import javax.swing.JFrame;
import javax.swing.JLabel;
import javax.swing.JPanel;
import javax.swing.JPasswordField;
import javax.swing.JTextField;

public class Dashboard extends JFrame implements ActionListener, KeyListener, MouseListener {

//JPanel
JPanel SignPane = new JPanel(new GridLayout(2,1));
JPanel LoginPane = new JPanel(new GridLayout(2,1));

//JLabel
JLabel gambar = new JLabel();
JLabel Userlbl = new JLabel("Username");
JLabel Passlbl = new JLabel("Password");

//TextField and Password
JTextField UserTxt = new JTextField(15);
JPasswordField PassTxt = new JPasswordField(15);

//JButton
JButton SigninBtn = new JButton("Sign In");
JButton SignupBtn = new JButton("Sign Up");
JButton SigninBtn2 = new JButton("Sign In");
JButton CancelBtn = new JButton("Cancel");

public Dashboard() {
    super("Boutique");

    gambar.setIcon(new ImageIcon("\\Untitled.jpg"));
    add(gambar, BorderLayout.NORTH);
    SignPane.add(SigninBtn);
    SignPane.add(SignupBtn);
    add(SignPane, BorderLayout.SOUTH);

    setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
    pack();
    setLocationRelativeTo(null);

    setResizable(false);
    SigninBtn.addMouseListener(this);
    SignupBtn.addMouseListener(this);
}

public void signin() {
    JFrame Signin = new JFrame();

    LoginPane.add(Userlbl);
    LoginPane.add(UserTxt);
    LoginPane.add(Passlbl);
    LoginPane.add(PassTxt);
    LoginPane.add(SigninBtn2);
    LoginPane.add(CancelBtn);
}

public static void main(String[] args) {
    EventQueue.invokeLater(new Runnable() {

        @Override
        public void run() {
            try {
                Dashboard window = new Dashboard();
                window.setVisible(true);
            } catch (Exception e){
                e.printStackTrace();
            }
        }
    });
}

@Override
public void mouseClicked(MouseEvent e) {
    if(e.getSource() == SigninBtn){
        signin();
    } else {
        dispose();
    }
}

@Override
public void mouseEntered(MouseEvent e) {
    // TODO Auto-generated method stub
}

@Override
public void mouseExited(MouseEvent e) {
    // TODO Auto-generated method stub
}

@Override
public void mousePressed(MouseEvent e) {
    // TODO Auto-generated method stub
}

@Override
public void mouseReleased(MouseEvent e) {
    // TODO Auto-generated method stub
}

@Override
public void keyPressed(KeyEvent e) {
    // TODO Auto-generated method stub
}

@Override
public void keyReleased(KeyEvent e) {
    // TODO Auto-generated method stub
}

@Override
public void keyTyped(KeyEvent e) {
    // TODO Auto-generated method stub
}

@Override
public void actionPerformed(ActionEvent e) {
    // TODO Auto-generated method stub
    }
}

Answer 1

我不明白你想要实现的目标。当你点击SigninBtn时，你想要关闭旧窗口并打开一个新窗口吗？

顺便说一句，使用actionPerformed方法而不是mouseClicked，如果你没有使用MouseListener的任何其他方法，那么将其删除。

SigninBtn.addActionListener(this);

@Override
public void actionPerformed(ActionEvent e){
    if(e.getSource().equals(SigninBtn)){
       signin();
    }
}

仪表板登录屏幕JAVA GUI

1 个答案: