提问者:小点点

Java . io . notserializableexception问题


我正在尝试发送以下内容,但收到以下错误。TableUser实现了可序列化,但问题似乎是FXCollections,但我不知道如何序列化。

这是Tableuser类。

package application;

import java.io.Serializable;


public class TableUser implements Serializable{

    private static final long serialVersionUID = 1L;
    private String username = "";

    public TableUser(String name) {
        this.username = name;
    }

    public String getUsername(){
        return username;
    }

    public void setUsername(String user){
        username = user;
    }

}






//NOT apart of TableUser - This is the code that isn't working
private static ObservableList<TableUser> clientList = FXCollections.observableArrayList(); 

Object[] data = new Object[2];
        data[0] = "CLIENTS";
        data[1] = clientList;

        for(int i = 0; i < clients.size(); i++){
            clients.get(i).sendData(data);
        }


//I don't know if this helps but here is the sendData method
protected void sendData(Object[] data){
        try {
            oos.writeObject(data); //ServerMultiClient.java:286
            oos.reset();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }


//This is from the Client application that is also part of the issue
if((fromServer = (Object[]) ois.readObject()) != null){ //Controller.java:109


java.io.WriteAbortedException: writing aborted; java.io.NotSerializableException: com.sun.javafx.collections.ObservableListWrapper
    at java.io.ObjectInputStream.readObject0(Unknown Source)
    at java.io.ObjectInputStream.readArray(Unknown Source)
    at java.io.ObjectInputStream.readObject0(Unknown Source)
    at java.io.ObjectInputStream.readObject(Unknown Source)
    at application.ChatRoomController$2.run(Controller.java:109)
    at java.lang.Thread.run(Unknown Source)
Caused by: java.io.NotSerializableException: com.sun.javafx.collections.ObservableListWrapper
    at java.io.ObjectOutputStream.writeObject0(Unknown Source)
    at java.io.ObjectOutputStream.writeArray(Unknown Source)
    at java.io.ObjectOutputStream.writeObject0(Unknown Source)
    at java.io.ObjectOutputStream.writeObject(Unknown Source)
    at application.ServerMultiClient.sendData(ServerMultiClient.java:286)
    at application.ServerMultiClient.run(ServerMultiClient.java:236)

共2个答案

匿名用户

鉴于< code > observablelist wrapper 是不可序列化的,您可以尝试循环访问< code>TableUser对象,并将它们逐个添加到< code>data中

大概是这样的:

Object[] data = new Object[clientList.size()+1];
data[0] = "CLIENTS";
int counter = 1;
for(TableUser tu: clientList) {
    data[counter] = tu;
    counter++;
}

匿名用户

当试图序列化的对象没有实现Seriezable类时,会抛出一个NotSeri的可序列化异常。在您的例子中,因为它告诉您的可序列化列表是不可序列化的,所以我推断clientList是不可序列化的。

您可以不使用Java FX < code > observable list ,而是使用一个列表,例如< code>java.util.ArrayList,它恰好实现了< code>Serializable。