java - JButton 不起作用？

Question

对于我正在制作的游戏，我制作了一个按钮类以用于菜单中的所有按钮。我终于设法让我的按钮出现在屏幕上，但现在当我单击它们时什么也没有发生。

我的按钮类:

package menu;

import java.awt.Image;
import java.awt.event.ActionEvent;
import java.awt.event.ActionListener;

import javax.swing.ImageIcon;
import javax.swing.JButton;
import javax.swing.JPanel;

public class Button extends JButton{

    public JButton button;
    public ImageIcon buttonImage;

    public int width, height;

    public String backgroundPath;
    public int x, y;
        public ActionListener listener;



public Button(String backgroundPath,int x, int y, ActionListener listener)

    {
        super();
        this.backgroundPath = backgroundPath;
        this.x = x;
        this.y = y;
        this.addActionListener(listener);   

        buttonImage = new 
            ImageIcon(PlayPanel.class.getResource(backgroundPath));
        button = new JButton();
        this.setIcon(buttonImage);
        this.setBounds(x, y, buttonImage.getIconWidth(), 
            buttonImage.getIconHeight());


    }


}

我知道我的菜单面板中可能有错误(@actionPerformed)。代码如下所示:

package menu;

import java.awt.*;
import java.awt.event.*;

import javax.swing.*;

@SuppressWarnings("serial")
public class MenuPanel extends JPanel implements ActionListener
{

    private Button playKnop, highScoreKnop, quitKnop, HTPKnop;
    private Tanks mainVenster;
    public MenuPanel menuPanel;

    int x = 95, width = 200, height = 50;



    public MenuPanel(Tanks mainVenster) 
    {
        this.mainVenster = mainVenster;
        this.setLayout(null); 

        playKnop = new Button("/buttons/PLAY.png",x, 350, menuPanel);       
        highScoreKnop = new Button("/buttons/HS.png",x, 460, menuPanel);
        HTPKnop = new Button("/buttons/HTP.png",x, 515, menuPanel);
        quitKnop = new Button("/buttons/QUIT.png",x, 570, menuPanel);


        this.add(playKnop);
        this.add(quitKnop);
        this.add(HTPKnop);
        this.add(highScoreKnop);

        validate();

    }


    public void actionPerformed(ActionEvent ae)
    {
        if (ae.getSource() == playKnop){
        mainVenster.switchPanel(new PlayPanel(mainVenster));

    } else if (ae.getSource() == quitKnop) {
        mainVenster.switchPanel(new QuitPanel(mainVenster));

    } else if (ae.getSource() == HTPKnop) {
        mainVenster.switchPanel(new HTPPanel(mainVenster));

    } else if (ae.getSource() == highScoreKnop) {
        mainVenster.switchPanel(new HSPanel(mainVenster));
    }

    }

}

我认为错误在于我正在编写 mainVenster(venster = 荷兰语面板/窗口)，我可能应该编写其他参数。问题是我不知道是哪些。

Answer 1

您正在监听器中定义按钮，因此您必须将类本身引用为 this:

playKnop = new Button("/buttons/PLAY.png",x, 350, this);       
highScoreKnop = new Button("/buttons/HS.png",x, 460, this);
HTPKnop = new Button("/buttons/HTP.png",x, 515, this);
quitKnop = new Button("/buttons/QUIT.png",x, 570, this);

您已经有一个 menuPanel 属性，因此在创建类时您可以分配一个值，在本例中为 this。

menuPanel = this;

但我不建议您删除此属性或将其重命名为 _this 或 _menuPanel 以同意约定。