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我希望将相同的 JMenuItems 添加到多个 JMenu,但它只显示最后一个 JMenu。这是我编写的代码。我希望为所有 JMenu 状态显示三个 JMenu 项目。使用此代码,前两个状态没有 JMenuItem,所有三个都只有最后一个。

import javax.swing.*;
import java.awt.event.*;  
public class Menu extends JFrame{
  public Menu()
  {
    super("Funky Menu");
JMenu [] states = new JMenu [3];
JMenuItem [] items = new JMenuItem [3];
//Initializing the items
items[0] = new JMenuItem("Industries");
items[0].setMnemonic('I');
items[1] = new JMenuItem("Hill Stations");
items[1].setMnemonic('H');
items[2] = new JMenuItem("Top Institutions");
items[2].setMnemonic('T');
//Initializing the states
//I've set the adjacent keys as the Mnemonics for easy user interaction
//though it is less intuitive, it can vary on the user preference.
states[0] =  new JMenu("Tamil Nadu"); states[0].setMnemonic('Q');
states[1] = new JMenu("West Bengal"); states[1].setMnemonic('W');
states[2] = new JMenu("Haryana"); states[2].setMnemonic('E');
//Adding all the items to each of the states
for(int i=0; i<3; ++i)
{
  for(int j=0; j<3; ++j)
  {
    states[i].add(items[j]);
  }
}
//adding action listener to menu items
for(int j=0; j<3; ++j)
{
  items[j].addActionListener(new ActionListener(){
    public void actionPerformed(ActionEvent evt)
  {
    //The next few lines could be clubbed together in one but for
    //clarity sake I write them seperately
    JMenuItem currentItem = (JMenuItem) evt.getSource();
    String textToDisplay = currentItem.getText();
    System.out.println(textToDisplay + " : located in ...");
    //one liner : System.out.println(((JMenuItem) evt.getSource()).getText() + " : located in ...");
  }
  });
}
//finally to fix up the MenuBar
JMenuBar bar = new JMenuBar();
setJMenuBar(bar);
for(int i=0; i<3; ++i)
{
  bar.add(states[i]);
}
getContentPane();
//TODO Create a JLabel add it to the contents
//Instead of writing to the console, update the frames text
setSize(500, 500);
setVisible(true);
 }

  public static void main(String[] args)
  {
    Menu app = new Menu();
    app.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
  }
}
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2 回答 2

4

我没有检查你的代码,但一个组件只能有一个父级。您需要创建单独的菜单项(可能使用相同的 Action 对象)。

于 2012-04-17T16:42:03.507 回答
2

@Puce 是正确的。相反,使用Action来封装共享功能,并使用Action. FileMenu是一个简单的例子。

于 2012-04-17T19:45:59.043 回答