你应该怎么做

Question

我知道这个问题在这里被无数次问了，我搜索了SO和其他来源的解决方案，但我无法解决错误。我有一个getter和setter用于我的ClientPlayer类属性，当调用某个按钮时，在GUI中调用setter，我想在客户端连接并将对象发送到服务器后使用getter。调用方法“this.client.sendTCP（clientPlayer.getPlayerName（））;”在ClientController中返回nullPointerException。

错误：“JavaFX Application Thread”java.lang.IllegalArgumentException：object不能为null。

我的猜测是我创建一个ClientPlayer的新实例一次太多，因此返回null，因为我的字符串playerName最初设置为none。但是，我不确定如何解决问题。我真的很感激任何帮助。

public class ClientPlayer implements Serializable {

public ClientPlayer() {

}

private String playerName;

public void setPlayerName(String playerName) {
    this.playerName = playerName;
   }

public String getPlayerName() {
    return this.playerName;
   }
}

在我的GUI代码的相关部分下面，我设置了一个连接服务器的按钮：

    ClientPlayer clientPlayer = new ClientPlayer();  

    clientToGame.setOnAction((ActionEvent w) -> {
         clientPlayer.setPlayerName(clientNameText.getText());
         clientController.connect();
         window.setScene(lobbyScene);
    });

以下是我的客户端类，我尝试获取名称并发送到服务器：

    public class ClientController() {
     ClientPlayer clientPlayer = new ClientPlayer();

    public void connect() {
    if (client.isConnected()) {
        Logger.getLogger(getClass().getName()).log(Level.INFO, "You are      already connected to :{0}", config.getHost());
        return;
    }
    this.client.start();
    try {
        this.client.connect(5000, config.getHost(), config.getTCPPort());
        System.out.println("Successfully connected to " + config.getHost());
        this.client.sendTCP(clientPlayer.getPlayerName());

    } catch (IOException ex) {
        Logger.getLogger(getClass().getName()).log(Level.SEVERE, "Server connection failed: {0}", ex.getMessage());
        throw new RuntimeException(ex);
    }

    MessageRegistry.registerMessages(client.getKryo());
    this.client.addListener(listener);
  }
 }

Answer 1

在GUI中，您创建一个ClientPlayer实例并在其上调用setPlayerName()。然后在ClientController中，您创建一个新的ClientPlayer实例（您永远不会在其上调用setPlayerName()），并在其上调用getPlayerName()。由于您从未为该实例设置播放器名称，因此getPlayerName()当然会返回null。

你需要决定自己的责任是什么？＃34; ClientPlayer实例。如果ClientController负责，则向getClientPlayer()添加ClientController方法，然后执行

clientToGame.setOnAction((ActionEvent w) -> {
     clientController.getClientPlayer().setPlayerName(clientNameText.getText());
     clientController.connect();
     window.setScene(lobbyScene);
});

并从GUI类中完全删除ClientPlayer。如果GUI类负责拥有它，那么将对它的引用传递给connect()方法，并从控制器中删除ClientPlayer字段：

public class ClientController() {
     // ClientPlayer clientPlayer = new ClientPlayer();

    public void connect(ClientPlayer clientPlayer) {
    if (client.isConnected()) {
        Logger.getLogger(getClass().getName()).log(Level.INFO, "You are      already connected to :{0}", config.getHost());
        return;
    }
    this.client.start();
    try {
        this.client.connect(5000, config.getHost(), config.getTCPPort());
        System.out.println("Successfully connected to " + config.getHost());
        this.client.sendTCP(clientPlayer.getPlayerName());

    } catch (IOException ex) {
        Logger.getLogger(getClass().getName()).log(Level.SEVERE, "Server connection failed: {0}", ex.getMessage());
        throw new RuntimeException(ex);
    }

    MessageRegistry.registerMessages(client.getKryo());
    this.client.addListener(listener);
  }
 }

当然

clientToGame.setOnAction((ActionEvent w) -> {
     clientPlayer.setPlayerName(clientNameText.getText());
     clientController.connect(clientPlayer);
     window.setScene(lobbyScene);
});

Answer 2

丑陋但应该有效：

使ClientPlayer访问静态。在你的gui班：

public static ClientPlayer clientPlayer = new ClientPlayer();

然后您可以通过以下方式访问ClientController中的同一个对象：

...
this.client.sendTCP(MyGuiClass.clientPlayer.getPlayerName());
...

编辑：问题是，你没有引用＆＃34;相同的＆＃34;您在ClientController中通过clientPlayer对象，就像在GUI-Class中一样。你应该以某种方式将该对象交给或使用静态引用。您也可以在ClientController的构造函数中交出它。

EDIT2：

你应该怎么做

ClientPlayer clientPlayer = new ClientPlayer();  

    clientToGame.setOnAction((ActionEvent w) -> {
         clientPlayer.setPlayerName(clientNameText.getText());
         clientController.connect(clientPlayer);
         window.setScene(lobbyScene);
    });

在您的ClientController中：

public void connect(ClientPlayer player) {
    this.clientPlayer = player;
    if (client.isConnected()) {
        Logger.getLogger(getClass().getName()).log(Level.INFO, "You are already connected to :{0}", config.getHost());
        return;
    }
    this.client.start();
    try {
        this.client.connect(5000, config.getHost(), config.getTCPPort());
        System.out.println("Successfully connected to " + config.getHost());
        this.client.sendTCP(player.getPlayerName());

    } catch (IOException ex) {
        Logger.getLogger(getClass().getName()).log(Level.SEVERE, "Server connection failed: {0}", ex.getMessage());
        throw new RuntimeException(ex);
    }

    MessageRegistry.registerMessages(client.getKryo());
    this.client.addListener(listener);
}

当我尝试发送到服务器时，为什么我的getter返回null？

2 个答案:

你应该怎么做