java - JPopupMenu 打开时开始鼠标拖动？

Question

我有一个弹出菜单，当鼠标悬停在组件上时会打开该菜单。现在我希望能够在弹出窗口仍然打开时开始拖动组件，但弹出菜单关闭事件总是消耗第一次鼠标单击。有解决方法吗？

Answer 1

     public class A
    .....
    implements MouseListener, ActionListener
      {
     JPopupMenu pmnu ;
    JMenuItem setcol ;

   JList lst ;
  Component cmp = null ;
 int x = 0, y = 0;
  ...
 private void
 pop ( MouseEvent ev )
 {
if ( ev.isPopupTrigger ()  )
 {
 cmp = ev.getComponent () ;
 x = ev.getX () ;
 y = ev.getY () ;
pmnu.show ( cmp, x, y ) ;
}
}
 public void mouseDragged( MouseEvent ev )
{
 pop ( ev ) ;
}
public void mouseReleased ( MouseEvent ev )
{
 pop ( ev ) ;
}
public void     actionPerformed( ActionEvent ev )
{
 Object src = ev.getSource() ;

 if ( src == setcol  &&  cmp != null )
 {
 cmp.setBackground ( Color.yellow ) ;
 return ;
}
}
public   A( )
{
pmnu = new JPopupMenu () ;

 setcol = new JMenuItem ( "Set color" ) ;

 pmnu.add ( setcol ) ;
....
}

public void
init()
{
 setcol.addActionListener ( this ) ;
....
 lst.addMouseListener ( this ) ;
 }
 public void mouseClicked(MouseEvent ev) {}
 public void mouseEntered(MouseEvent ev) {}
 public void mouseExited(MouseEvent ev) {}
}