我正在用java编程一个在线视频游戏。我已经完成了服务器,现在我进入了客户端。 我的问题在于socket侦听器代码中的某个地方,这是一个swingworker子类,它的任务是侦听服务器(doInBackGround())并根据需要更新游戏地图。Java Swing中的差异发布()/ process()交互
下面的代码:
import javax.swing.*;
import java.util.List;
public class GameWorker extends SwingWorker<Void, String> {
private SocketStreamsBean streams;
private GameFrame game;
public GameWorker(SocketStreamsBean streams, GameFrame game) {
this.streams = streams;
this.game = game;
}
@Override
protected Void doInBackground() throws Exception {
for(String msg = streams.getIn().readLine(); msg != null; msg = streams.getIn().readLine()){
System.out.println("bp " + msg + " " + Thread.currentThread().getId());//TODO remove
publish(msg);
System.out.println("ap " + msg + " " + Thread.currentThread().getId());//TODO remove
}
return null;
}
@Override
protected void process(List<String> list) {
for(String msg = list.remove(0); list.size() != 0; msg = list.remove(0)) {
System.out.println("dp " + msg + " " + Thread.currentThread().getId());//TODO remove
String[] cmds = msg.split(":");
switch (cmds[0]) {
case "ADD":
game.add(cmds[1], cmds[2], cmds[3]);
break;
case "MOVE":
game.remove(cmds[1]);
game.add(cmds[1], cmds[2], cmds[3]);
break;
case "REMOVE":
game.remove(cmds[1]);
break;
case "BULLETS":
//game.addBullets(cmds[1]);
}
}
list.clear();
}
}
据三米调试的println()当玩家移动和服务器广播它的信息被读取并公布,但从来没有处理的所有客户端。怎么会这样?
不要从内部列表中删除过程方法。简单地迭代和处理。 –