小编Cha*_*had的帖子

我遵循了在java中创建迷宫游戏的教程.完成后,我决定添加一些东西(主菜单,更多级别等).我目前卡在主菜单上,因为我无法让鼠标输入工作.当我在开始游戏框中单击鼠标时,我试图切换到屏幕= 1.这是我目前的课程代码.

package Maze;

import java.awt.Color;
import java.awt.Font;
import java.awt.Graphics;
import java.awt.Rectangle;
import java.awt.event.ActionEvent;
import java.awt.event.ActionListener;
import java.awt.event.KeyAdapter;
import java.awt.event.KeyEvent;
import java.awt.event.MouseAdapter;
import java.awt.event.MouseEvent;
import java.awt.event.MouseListener;

import javax.swing.JPanel;
import javax.swing.SwingUtilities;
import javax.swing.Timer;

public class Board extends JPanel implements ActionListener{
    private Timer timer;    
    private Map m;
    private Player p;
    private int screen = 0;
    private String Message = "";
    private Font font = new Font("Serif", Font.BOLD, 48);
    private Font start = new Font("Serif", Font.BOLD, 36);
    private Rectangle start2 = new Rectangle(135,350,180,55); …

我试图在单击标题时使用setValueAt方法在我的jtablemodel中设置一个值.我目前有一个MouseAdapter正确显示正在单击哪个列但是没有调用board.setValueAt()方法.

鼠标监听器.这会正确显示我单击的列.我试过使用board.getModel().setValueAt()但是也没用.

public class HeaderListener extends MouseAdapter{
    private JTable board;
    public HeaderListener(JTable board){
        this.board = board;
    }

    @Override
    public void mouseClicked(MouseEvent event) {
        Point point = event.getPoint();
        int column = board.columnAtPoint(point);
        System.out.println(column);
        Data value = new Data(0,"yes");
        board.setValueAt(value, 0, column);
    }
}

TableModel."输入时设置值"永远不会显示给控制台.当我尝试覆盖它时,它给了我一个错误.

public class TableModel extends AbstractTableModel{
    private Data[][] data; 
    private String[] columnNames = {"Move Here",
                        "Move Here",
                        "Move Here",
                        "Move Here",
                        "Move Here","Move Here"};

    public TableModel(){
        data = new Data[6][7];
        Data test = new Data(0,"yes");
        data[1][1] …