Question

所以我试图将一个Object从客户端发送到服务器，然后我用这个对象做一些事情。我确认在发送时对象包含数据。当我在调试器中读取我的对象时，它的所有属性都是null。

以下是我发送对象的地方：

private class registerUser extends Task<Void>
{
        Users user;
        private static final String code = ".register";
        private static final String host = "localhost";
        private static final int portNumber = 4444;
        private clientThread clientThread;


    public registerUser(Users user)
    {
        this.user = user;
    }

    @Override
    protected Void call() throws Exception
    {
        try
        {
            Socket socket = new Socket(host, portNumber);
            Thread.sleep(1000);

            //Setup I/O
            ObjectOutputStream outToServerObject = new ObjectOutputStream(socket.getOutputStream());
            //PrintWriter serverOutString = new PrintWriter(socket.getOutputStream(), false);
            InputStream serverInString = socket.getInputStream();

            //serverOutString.println(code);
           // serverOutString.flush();

            outToServerObject.writeObject(user);
            outToServerObject.flush();
        }
        catch (IOException e)
        {
            System.err.println("Fatal Connection error!");
            e.printStackTrace();
        }
        return null;
    }

当我按下GUI中的按钮时，将使用上述任务，如下所示：

public void btn_registerUser(ActionEvent actionEvent) throws IOException
{

        user.setUserName(txt_Username.getText());
        user.setFirstName(txt_FirstName.getText());
        user.setLastName(txt_LastName.getText());
        user.setCity(txt_City.getText());
        user.setBirthday(date_Birthday.getValue());


        Task<Void> task = new registerUser(user);
        Thread thread = new Thread(task);
        thread.setDaemon(true);
        thread.start();

}

收到对象的服务器中的代码：

//Setup I/O
this.clientOut = new PrintWriter(socket.getOutputStream(), false);
ObjectInputStream inFromClientObject = new ObjectInputStream(socket.getInputStream());


// Scanner in = new Scanner(socket.getInputStream());
Users user = (Users) inFromClientObject.readObject();
doSomething(user);

User.java类，它是辩论中的对象

public class Users implements Serializable
{

    /*
    Variables
    TODO Create a friend list
    */
    private transient IntegerProperty userID;
    private transient StringProperty userName;
    private transient StringProperty firstName;
    private transient StringProperty lastName;
    private transient StringProperty city;
    private transient ObjectProperty<LocalDate> birthday;
    private transient ListProperty<String> musicGenre;
    private transient List<String> musicGenres = new ArrayList<>();
    /**
     * Default constructor.
     */
    public Users()
    {
        this(null,null,null,null,null);
    }

    /**
     * Constructor with some initial data.
     * Takes in a {@code String} and sets a reference
     *
     * @param userName
     *
     */
    public Users(String userName,  String firstName, String lastName, String city, LocalDate birthday)
    {
        this.userName = new SimpleStringProperty(userName);
        this.userID = new SimpleIntegerProperty();
        this.firstName = new SimpleStringProperty(firstName);
        this.lastName = new SimpleStringProperty(lastName);
        this.city = new SimpleStringProperty(city);
        this.birthday = new SimpleObjectProperty<LocalDate>(birthday);
        ObservableList<String> observableList = FXCollections.observableArrayList(this.musicGenres);
        this.musicGenre = new SimpleListProperty<String>(observableList);

    }
}

由于这个问题已经很久了，我已经把吸气剂和制定者排除在外了

因此，在运行服务器的调试器中，当我doSomething(user)

时，用户的所有属性都为空

*修改这些建议正在制定中，但我遇到了问题

private List<String> musicGenreList = new ArrayList<>();
ObservableList<String> observableList = FXCollections.observableArrayList(musicGenreList);
private ListProperty<String> musicGenre = new SimpleListProperty<String>(observableList);

这是我可以向ListProperty添加项目的唯一方法，但observableList无法序列化

Answer 1

像James_D一样，基本上我必须创建自定义WriteExternal和ReadExternal

在Users.java中需要进行更改

public class Users implements Externalizable
{
    public Users(String userName,  String firstName, String lastName, String city, LocalDate birthday)
    {
        setUserName(userName);
        setFirstName(firstName);
        setBirthday(birthday);
    }

    private IntegerProperty userID = new SimpleIntegerProperty(this, "id");
    private StringProperty userName = new SimpleStringProperty(this, "userName');
    private ObjectProperty<LocalDate> birthday = new SimpleObjectProperty<>(this, "birthday");

    //Create setters and getters...

    @Override
    public void writeExternal(ObjectOutput out) throws IOException
    {
        out.writeInt(getUserID());
        out.writeObject(getUserName());
        out.writeObject(getBirthday());
    }

    @Override
    public void readExternal(ObjectInput in) throws IOException, ClassNotFoundException
    {
        setUserID((int) in.readInteger());
        setUserName((String) in.readObject());
        setBirthday((LocalDate) in.readObject());
    }
}

<强>然而我还没有能够使用以下方法序列化ListProperty

private ObservableList<String> observableList = FXCollections.observableArrayList();
private ListProperty<String> newList = new SimpleListProperty<>(this, "musicGenre", observableList);

根据我的需要，需要以这种方式实例化ListProperty，因为我需要使用user.newListProperty().get().add("String")向其添加String项而不使用ObservableList我无法执行此操作

如果有人知道如何解决这个问题，我们将不胜感激

ObjectInputStream返回一个空对象

1 个答案: