小编Sav*_*orm的帖子

此代码应在输入对话框中接收全名字符串示例"Billy Bob Smith",并在消息对话框中将首字母输出为字母组合示例"BBS"​​.但由于某种原因,main方法不允许我访问fullName变量.

import javax.swing.*;

public class HardMonogram {
     //---------- ATTRIBUTES ----------//
     private String fullName;
     private String monogram;
     private String first;
     private String middle;
     private String last;


     //---------- METHODS ----------//
     public String getInitial(String seperateName) {
           return seperateName.substring(0, 1);
     }

     public void getSeperateName(String fullName) {
           first  = fullName.substring(0, fullName.indexOf(" "));
           middle = fullName.substring(fullName.indexOf(" ") + 1, fullName.length());
           last   = middle.substring(middle.indexOf(" ") + 1, middle.length());
           middle = middle.substring(0, middle.indexOf(" "));
     }

    public void setMonogram() {
          monogram = getInitial(first)  +
                     getInitial(middle) +
                     getInitial(last); …

这段代码构建正确,一切似乎都有效,但键无效.我认为它是动作监听器或椭圆形不更新.我正在尝试通过初学者java游戏编程.我相信这很容易,但我没有抓住它.如果这有所不同,我在sublime text 2中使用mac.

package javagame;

import javax.swing.*;
import java.awt.*;
import java.awt.event.*;

public class JavaGame extends JFrame {

int x, y;

public class AL extends KeyAdapter {

    public void keyPressed(KeyEvent e) {
        int keycode = e.getKeyCode();
        if(keycode == e.VK_LEFT); {
            x-= 3;
        }
        if(keycode == e.VK_RIGHT); {
            x+= 3;
        }
        if(keycode == e.VK_UP); {
            y-= 3;
        }
        if(keycode == e.VK_DOWN); {
            y+= 3;
        }
    }


    public void keyReleased(KeyEvent e) {

    }
}

public JavaGame() {
    addKeyListener(new AL());
    setTitle("Jave Game");
    setSize(700, 700); …