微信公众号搜"智元新知"关注
微信扫一扫可直接关注哦!

在Swing中实现后退/前进按钮

如何解决在Swing中实现后退/前进按钮

| 我有一个快速的问题。 我对Swing有了一点经验,最简单的方法是绘制一个相当大的GUI。 作为GUI的一部分,我想要前进和后退按钮。我正在尝试采用的方法是实现将当前JPanel推入堆栈并检索先前值(正向或反向(因此为2个堆栈))的方法。但是我无法正常工作。也许我正在完全以错误的方式进行操作,或者可能无法在使用它的方式中使用堆栈。无论哪种情况,它真的让我烦恼。我想可能有更简单的方法,例如卡片布局,但我认为这种方法应该可行,这真是令人讨厌。 可能值得注意的是,我正在使用JFrame \“ base class \\”并根据屏幕更改中央JPanel。导航栏作为“基类”的一部分是恒定的,但是 此“基类”的代码
public class Main_Frame extends JFrame{
    static JPanel nav_bar_panel;
    JButton home;
    JButton back;
    JButton forward;
    JPanel currentPanel;

    static Stack<JPanel> prevIoUsPanels;
    static Stack<JPanel> forwardPanels;

    public Main_Frame(){
        super(\"DEMO\");
        setSize(800,600);
        setLayout(new BorderLayout());
        setVisible(true);

        add(nav_bar(),BorderLayout.norTH);
        currentPanel = init_display();
        add(currentPanel,BorderLayout.CENTER);

        prevIoUsPanels = new Stack<JPanel>();
        forwardPanels  = new Stack<JPanel>();
    }

    private JPanel nav_bar(){
        ButtonPressHandler handler = new ButtonPressHandler();

        nav_bar_panel = new JPanel(new FlowLayout(FlowLayout.LEFT,10,10));
        back = new JButton(\"Back\");
        back.addActionListener(handler);
        home = new JButton(\"Home\");
        home.addActionListener(handler);
        forward = new JButton(\"Forward\");
        forward.addActionListener(handler);

        nav_bar_panel.add(back);
        nav_bar_panel.add(home);
        nav_bar_panel.add(forward);

        return nav_bar_panel;
    }

    private JPanel init_display(){
        Home_Panel home_panel = new Home_Panel();

        return home_panel;
    }

    public void change_display(JPanel myPanel){
        invalidate();
        remove(currentPanel);
        prevIoUsPanels.push(currentPanel);
        currentPanel = myPanel;
        add(currentPanel);
        validate();
    }

    public void prevIoUs_display(){
        if(!prevIoUsPanels.empty()){
            invalidate();
            remove(currentPanel);
            forwardPanels.push(currentPanel);
            currentPanel = prevIoUsPanels.pop();
            add(currentPanel);
            validate();
        }
    }

    public void forward_display(){
        if(!forwardPanels.empty()){
            invalidate();
            remove(currentPanel);
            prevIoUsPanels.push(currentPanel);
            currentPanel = forwardPanels.pop();
            add(currentPanel);
            validate();
        }
    }

    private class ButtonPressHandler implements ActionListener 
       {
          public void actionPerformed( ActionEvent event )
          {
              if(event.getSource() == back){
                  prevIoUs_display();
                  System.out.print(\"You selected back\");
              } else if(event.getSource() == forward){
                  forward_display();
                  System.out.print(\"You selected forward\");
              }
          } // end method actionPerformed
       } // end private inner class TextFieldHandler

}
    

解决方法

这是使用ѭ1的示例。
import java.awt.BorderLayout;
import java.awt.CardLayout;
import java.awt.Color;
import java.awt.Dimension;
import java.awt.EventQueue;
import java.awt.event.ActionEvent;
import java.util.Random;
import javax.swing.AbstractAction;
import javax.swing.JButton;
import javax.swing.JFrame;
import javax.swing.JLabel;
import javax.swing.JPanel;

/** @see http://stackoverflow.com/questions/5654926 */
public class CardPanel extends JPanel {

    private static final Random random = new Random();
    private static final JPanel cards = new JPanel(new CardLayout());
    private final String name;

    public CardPanel(String name) {
        this.name = name;
        this.setPreferredSize(new Dimension(320,240));
        this.setBackground(new Color(random.nextInt()));
        this.add(new JLabel(name));
    }

    @Override
    public String toString() {
        return name;
    }

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

            @Override
            public void run() {
                create();
            }
        });
    }

    private static void create() {
        JFrame f = new JFrame();
        f.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
        for (int i = 1; i < 9; i++) {
            CardPanel p = new CardPanel(\"Panel \" + String.valueOf(i));
            cards.add(p,p.toString());
        }
        JPanel control = new JPanel();
        control.add(new JButton(new AbstractAction(\"\\u22b2Prev\") {

            @Override
            public void actionPerformed(ActionEvent e) {
                CardLayout cl = (CardLayout) cards.getLayout();
                cl.previous(cards);
            }
        }));
        control.add(new JButton(new AbstractAction(\"Next\\u22b3\") {

            @Override
            public void actionPerformed(ActionEvent e) {
                CardLayout cl = (CardLayout) cards.getLayout();
                cl.next(cards);
            }
        }));
        f.add(cards,BorderLayout.CENTER);
        f.add(control,BorderLayout.SOUTH);
        f.pack();
        f.setLocationRelativeTo(null);
        f.setVisible(true);
    }
}
    ,  使我可以重用的想法很不错。可惜秋千没有内置此功能 查看Card Layout Actions(卡布局操作),这些操作可能会尝试使Card Layout变得更易于使用。     ,我通常的做法如下: 我有一个StepManager类(编写一次,永久使用),该类处理与步骤相关的所有逻辑。它具有next(),previous(),reset(),isFirst()和isLast()之类的方法。 然后,我会得到带有适当操作(或您选择用于侦听用户交互的任何内容)的“下一步”和“上一步”按钮。 与\'Next \'按钮相关的代码调用stepManager.next()来检索下一步的索引。然后(当我完成下一步时)我只需调用(另一种方法)showStep(int index)以显示与当前步骤索引相对应的实际步骤用户界面。 每个步骤都是一个单独的JPanel(Step01,Step02,Step03 ...)。
public void showStep(int index) {
    ContentPanel.removeAll();
    ContentPanel.setLayout(new BorderLayout());

    switch (index) {
        case 0:
            ContentPanel.add(Step01,BorderLayout.CENTER);
            break;

        case 1:
            ContentPanel.add(Step02,BorderLayout.CENTER);
            break;

        case 2:
            ContentPanel.add(Step03,BorderLayout.CENTER);
            break;

        case 3:
            ContentPanel.add(Step04,BorderLayout.CENTER);

        }

    ContentPanel.validate();
    ContentPanel.repaint();
}
    

版权声明:本文内容由互联网用户自发贡献,该文观点与技术仅代表作者本人。本站仅提供信息存储空间服务,不拥有所有权,不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容, 请发送邮件至 dio@foxmail.com 举报,一经查实,本站将立刻删除。