Java 按下按钮后,打开另一个面板内的另一个面板
我得到一个错误,说我没有添加一些方法(执行的操作),但我已经添加了。我打开配电盘有困难Java 按下按钮后,打开另一个面板内的另一个面板,java,user-interface,jpanel,action,panel,Java,User Interface,Jpanel,Action,Panel,我得到一个错误,说我没有添加一些方法(执行的操作),但我已经添加了。我打开配电盘有困难 public class panel1 extends JPanel implements ActionListener(){ private panel2 p2=new panel2(); private JButton button; public panel1(){ button=new JButton("open panel2"); add(button,Bo
public class panel1 extends JPanel implements ActionListener(){
private panel2 p2=new panel2();
private JButton button;
public panel1(){
button=new JButton("open panel2");
add(button,BorderLayout.BEFORE_FIRST_LINE);
button.addActionListener(new ActionListener(){
@Override
public void actionPerformed(ActionEvent ae) {
add(p2);
}
});
}
}考虑这些变化:
public class panel1 extends JPanel implements ActionListener(){
private panel2 p2=new panel2();
private JButton button;
public panel1(){
button=new JButton("open panel2");
add(button,BorderLayout.BEFORE_FIRST_LINE);
button.addActionListener(new ActionListener(){
@Override
public void actionPerformed(ActionEvent ae) {
//Add readability: Where to add?
panel1.this.add(p2);
}
});
}
//THIS HERE makes your panel1 implment ActionListener
@Override
public void actionPerformed(ActionEvent ae) {
}
}
还请注意常见的Java命名约定-类名应以大写字母开头(
Panel1扩展了JPanel
)Panel1需要实现ActionListener接口所描述的actionPerformed方法,但还不是已经具有@Override public void actionPerformed(ActionEvent ae)的方法{add(p2);}?很抱歉,我是新来编程的。不,这是匿名类的实现。谢谢!这消除了错误。虽然它仍然没有打开第二个面板。也许你应该添加(panel2,BorderLayout.CENTER)或其他什么?
You can check the below code for replacing jpanel without switching jframe.
contentPanel.removeAll();
contentPanel.repaint();
contentPanel.revalidate();
contentPanel.add(//add your panel here);
contentPanel.repaint();
contentPanel.revalidate();