当我尝试发送到服务器时，为什么我的获得者返回null？

Question

我知道这个问题已经在这里被问过无数次了，我已经搜索过SO和其他来源的解决方案，但我无法解决这个错误。我为我的ClientPlayer类属性设置了一个getter和setter，当调用某个按钮时，setter在GUI中调用，并且在客户端连接并将对象发送到服务器后，我想使用getter。调用方法“this.client.sendTCP（clientPlayer.getPlayerName（））;”在ClientController中返回一个nullPointerException。

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

我的猜测是我创建一个新的ClientPlayer实例太多，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。我真的很感谢任何帮助。

public class ClientPlayer implements Serializable { 

public ClientPlayer() { 

} 

private String playerName; 

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

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

低于我的GUI代码，我设置一个按钮连接到服务器的相关部分：

ClientPlayer clientPlayer = new ClientPlayer(); 

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

这里是我的客户级的位，我试图让命名并发送至服务器：

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

在你的GUI，您可以创建一个ClientPlayer实例，并调用它setPlayerName()。然后在ClientController中，创建一个新的ClientPlayer实例（您永远不会呼叫setPlayerName()），并调用getPlayerName()。由于您从未为该实例设置了玩家名称，因此getPlayerName()当然会返回null。

您需要决定自己的责任是“拥有”ClientPlayer实例。如果是ClientController的责任，那么无论是getClientPlayer()方法添加到ClientController，并做

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

和完全来自你的GUI类中删除ClientPlayer。如果是GUI类拥有它的责任，然后传递一个参考给它的connect()方法，并从控制器移除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); 
    } 
} 

当然和

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

Answer 2

丑，但应该工作：

充分利用ClientPlayer访问静态。在您的GUI类：通过clientPlayer 的事情是，你不引用“相同”的对象：

public static ClientPlayer clientPlayer = new ClientPlayer(); 

，那么你可以访问相同的对象在ClientController通过：

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

编辑在你的ClientController中，就像在GUI-Class中一样。您应该以某种方式将该对象移交或使用静态引用。你也可以在ClientController的构造函数中交给它。

EDIT2：

如何你应该这样做

ClientPlayer clientPlayer = new ClientPlayer(); 

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

在你ClientController：

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); 
}