我正在为我的网络课制作多人冒险游戏。我有一个客户端和一个服务器，服务器是多线程的，每当它连接一个新客户端时就会启动一个新线程。我有一个数组列表，跟踪球员，以确保没有添加新的球员。出于某种原因，当新客户端连接时，它取代旧客户端并填充新位置。这是我的这部分代码
public class ClientHandler implements Runnable{
private AsynchronousSocketChannel clientChannel;
private static String command[];
private static String name;
private static GameCharacter character;
public ClientHandler(AsynchronousSocketChannel clientChannel)
{
this.clientChannel = clientChannel;
}
public void run(){
try{
System.out.println("Client Handler started for " + this.clientChannel);
System.out.println("Messages from Client: ");
while ((clientChannel != null) && clientChannel.isOpen()) {
ByteBuffer buffer = ByteBuffer.allocate(32);
Future result = clientChannel.read(buffer);
//Wait until buffer is ready
result.get();
buffer.flip();
String message = new String(buffer.array()).trim();
if(message == null || message.equals(""))
{
break;
}
System.out.println(message);
clientChannel.write(buffer);
try {
//Add the character to the routing table and the character table
if (message.contains("connect")) {
System.out.println("I'm here too?");
command = message.split(" ");
name = command[1];
AdventureServer.userInfo.put(name, this);
//Check to see if this game character exists
GameCharacter test;
boolean exists = false;
for(int i=0; i < AdventureServer.characters.size(); i++)
{
test = AdventureServer.characters.get(i);
System.out.println(test.getName());
System.out.println(this.name);
if(this.name.equals(test.getName()))
{
System.out.println("already Here");
exists = true;
}
}
if (exists == true)
{
//This person has connected to the server before
}
else {
//Create a game character
System.out.println("didn't exist before");
character = new GameCharacter(this.name, World.getRow(), World.getCol());
AdventureServer.characters.add(AdventureServer.userInfo.size() - 1, character);
System.out.println(AdventureServer.characters.get(0).getName() + " " +AdventureServer.characters.get(1).getName());
}
}
据我所知，底部的打印行会为连接的第一个客户端引发错误，但这不是问题的一部分。 这是服务器的声明
public class AdventureServer {
public static Map<String, ClientHandler> userInfo = new HashMap<>();
public static World world;
public static List<GameCharacter> characters = Collections.synchronizedList(new ArrayList<>());
public static void main(String args[]) {
//Create the games map that all of the users will exist on
world = new World(args[0]);
System.out.println("Asynchronous Chat Server Started");
try {
AsynchronousServerSocketChannel serverChannel = AsynchronousServerSocketChannel.open();
InetSocketAddress hostAddress = new InetSocketAddress("192.168.1.7", 5000);
serverChannel.bind(hostAddress);
while (true)
{
System.out.println("Waiting for client to connect");
Future acceptResult = serverChannel.accept();
AsynchronousSocketChannel clientChannel = (AsynchronousSocketChannel) acceptResult.get();
new Thread (new ClientHandler(clientChannel)).start();
}
} catch (Exception e) {
System.out.println("error interrupted");
e.printStackTrace();
System.exit(0);
}
}
}
这是我的游戏角色构造函数
public class GameCharacter {
public static int xpos;
public static int ypos;
private static String name;
private static int rowSize;
private static int columnSize;
static List<String> inventory = new ArrayList<>();
//Constructor
GameCharacter(String n, int rSize, int cSize)
{
xpos = 0;
ypos = 0;
name = n;
rowSize = rSize;
columnSize = cSize;
}
GameCharacter()
{
xpos = 0;
ypos = 0;
name = "billybob";
rowSize = 10;
columnSize = 10;
}
答案 0 :(得分：0)
作为可读性，可测试性和样式的问题，我还建议您不要直接访问属于另一个类的数据结构。而不是
Adventureserver.characters.add(blah blah)
我建议将字符设置为Adventureserver的私有字段，然后创建一个方法来添加或删除字符。事实上，我倾向于不让角色静态 - 没有真正的优势，你可能在某些时候想要运行多个Adventureserver。
有点像：
public class AdventureServer {
<...>
private List<GameCharacter> characters = Collections.synchronizedList(new ArrayList<>);
<...>
public void addCharacter(GameCharacter char) {
<... error checking ...>
characters.add(char);
}
public void removeCharacter(GameCharacter char) {
<... implementation ... >
}
public boolean isCharacterHere(GameCharacter char) {
}
public List<GameCharacter> getCharacters() {
<... you could either return characters here, or a copy of it,
depending upon how paranoid you want to be >
答案 1 :(得分：0)
您可以尝试：
public static volatile List<GameCharacter> characters = Collections.synchronizedList(new ArrayList<>());
更新： 问题是您使用的是非同步的HashMap userInfo。 从以下地址更改该行：
AdventureServer.characters.add(AdventureServer.userInfo.size() - 1, character);
要：
AdventureServer.characters.add(character);
或者让您的HashMap同步：
public static Map<String, ClientHandler> userInfo = Collections.synchronizedMap(new HashMap<>());
所有这些静态声明都会出现问题，您应该删除它们。通常，您应该避免使用静态。
ClientHandler的：
private static String command[];
private static String name;
private static GameCharacter character;
GameCharacter：
public static int xpos;
public static int ypos;
private static String name;
private static int rowSize;
private static int columnSize;
static List<String> inventory = new ArrayList<>();
只是旁注，这样你的类更像Java代码应该是这样的：
import java.util.ArrayList;
import java.util.List;
public class GameCharacter {
private int xpos;
private int ypos;
private String name;
private int rowSize;
private int columnSize;
private List<String> inventory = new ArrayList<>();
// Constructor
GameCharacter(String n, int rSize, int cSize) {
this.xpos = 0;
this.ypos = 0;
this.name = n;
this.rowSize = rSize;
this.columnSize = cSize;
}
GameCharacter() {
this.xpos = 0;
this.ypos = 0;
this.name = "billybob";
this.rowSize = 10;
this.columnSize = 10;
}
public int getXpos() {
return xpos;
}
public void setXpos(int xpos) {
this.xpos = xpos;
}
public int getYpos() {
return ypos;
}
public void setYpos(int ypos) {
this.ypos = ypos;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public int getRowSize() {
return rowSize;
}
public void setRowSize(int rowSize) {
this.rowSize = rowSize;
}
public int getColumnSize() {
return columnSize;
}
public void setColumnSize(int columnSize) {
this.columnSize = columnSize;
}
public List<String> getInventory() {
return inventory;
}
public void setInventory(List<String> inventory) {
this.inventory = inventory;
}
}