Question

我正在用Java编写Risk的克隆版，我的代码遇到了一些麻烦。当我创建一个新游戏时，我创建了一个JPanel，其中包含一个JTextField（用于玩家名称）和一个JComboBox（用于玩家颜色），一个面板供用户想要创建的每个玩家使用。该面板的实例是基于第二个JComboBox动态创建的，该第二个JComboBox允许用户从3到8中选择多个玩家。

我的问题是，当我想从输入到玩家创建面板的数据创建玩家对象时，每个玩家对象都会从最近创建的玩家创建面板中检索数据。我有一个功能性的解决方案，但我似乎无法弄清楚为什么似乎是“正确的”解决方案（）将无法正常工作。

这是我现在正在使用的代码：创建面板类：

package risk;

import java.awt.Color;
import java.awt.Dimension;
import java.awt.event.ActionEvent;
import java.awt.event.ActionListener;

import javax.swing.ImageIcon;
import javax.swing.JComboBox;
import javax.swing.JLabel;
import javax.swing.JPanel;
import javax.swing.JTextField;

public class PlayerCreatePanel extends JPanel{

public static ImageIcon[] playerColors = {Resources.red,Resources.green,Resources.blue,Resources.cyan,Resources.magenta,Resources.yellow,Resources.orange,Resources.gray};

public JComboBox playerColor;

public JTextField playerName;

public PlayerCreatePanel(int index){

    this.setPreferredSize(new Dimension(360, 30));

    JLabel numberLabel = new JLabel ("Player " + index + ":  ");
    JLabel nameLabel = new JLabel("Name: ");
    JLabel colorLabel = new JLabel("  Color: ");

    playerName = new JTextField("");
    playerName.setColumns(13);

    playerColor = new JComboBox(playerColors);

    this.add(numberLabel);
    this.add(nameLabel);
    this.add(playerName);
    this.add(colorLabel);
    this.add(playerColor);
}
}

对于新游戏类，创建玩家的部分：

package risk;

import java.awt.Color;
import java.awt.Dimension;
import javax.swing.JPanel;

public class NewGame {

private static PlayerCreatePanel [] panels = {  new PlayerCreatePanel(1), new PlayerCreatePanel(2), 
                                                new PlayerCreatePanel(3), null, null, null, null, null};

private static int playerCount = 3;

public static void createPlayers(){
    Resources.players = new Player[playerCount];
    switch(playerCount){
    case 8: Resources.players[7] = new Player (panels[7].playerName.getText(), Resources.colors[panels[7].playerColor.getSelectedIndex()], 8);
    case 7: Resources.players[6] = new Player (panels[6].playerName.getText(), Resources.colors[panels[6].playerColor.getSelectedIndex()], 7);
    case 6: Resources.players[5] = new Player (panels[5].playerName.getText(), Resources.colors[panels[5].playerColor.getSelectedIndex()], 6);
    case 5: Resources.players[4] = new Player (panels[4].playerName.getText(), Resources.colors[panels[4].playerColor.getSelectedIndex()], 5);
    case 4: Resources.players[3] = new Player (panels[3].playerName.getText(), Resources.colors[panels[3].playerColor.getSelectedIndex()], 4);
    case 3: Resources.players[2] = new Player (panels[2].playerName.getText(), Resources.colors[panels[2].playerColor.getSelectedIndex()], 3);
            Resources.players[1] = new Player (panels[1].playerName.getText(), Resources.colors[panels[1].playerColor.getSelectedIndex()], 2);
            Resources.players[0] = new Player (panels[0].playerName.getText(), Resources.colors[panels[0].playerColor.getSelectedIndex()], 1); break;
    default: break;     
    }
}
}

现在，我所学到的是正确的方法是使我的JTextField和JComboBox成为私有，并编写访问器，如下所示：

private JComboBox playerColor;
private JTextField playerName;

//...same method as above

public static String getName(){
    return playerName.getText();
}

public static int getColorIndex(){
    return playerColor.getSelectedIndex();
}

并更改新游戏方法中的创建行以阅读如下内容：

Resources.players [0] = new Player (panels[0].getName(), panels[0].getColorIndex(), 1);

扩展JPanel的类中的访问器

0 个答案: