Java 使用ObjectOutputStream保存HashMap

Java 使用ObjectOutputStream保存HashMap,java,serialization,objectoutputstream,bukkit,Java,Serialization,Objectoutputstream,Bukkit,为了在命令中保存远程传送点,我有一个哈希映射: public HashMap<Player, Location> mapHomes = new HashMap<>(); 由于重新启动时应保持这些设置,因此我实现了一种保存方法: public void save(HashMap<Player,Location> mapHome, String path) throws NotSerializableException{ try{ Obj

为了在命令中保存远程传送点,我有一个
哈希映射

public HashMap<Player, Location> mapHomes = new HashMap<>();
由于重新启动时应保持这些设置,因此我实现了一种保存方法:

public void save(HashMap<Player,Location> mapHome, String path) throws NotSerializableException{
    try{
        ObjectOutputStream oos = new ObjectOutputStream(new FileOutputStream(path));
        oos.writeObject(mapHome);
        oos.flush();
        oos.close();
    }catch(Exception e){
        e.printStackTrace();
    }
}
public void save(HashMap mapHome,字符串路径)引发NotSerializableException{
试一试{
ObjectOutputStream oos=新的ObjectOutputStream(新文件输出流(路径));
oos.writeObject(mapHome);
oos.flush();
oos.close();
}捕获(例外e){
e、 printStackTrace();
}
}
但它不起作用。它抛出
NotSerializableException

我认为主要的问题是
Player
Location
不是可序列化的类型,那么我应该如何编写这个
HashMap

class Player implements Serializable {}

class Location implements Serializable {}
请记住,您只能序列化实现
Serializable
接口的对象。
因此,您的
Player
Location
类也必须实现该接口。

我认为您正在将
HashMap
写入文件

您需要做的是使您的
播放器
位置
类可序列化

public class Player implements java.io.Serializable {
    // ...
}

public class Location implements java.io.Serializable {
    // ...
}

已可序列化。

HashMap
已可
序列化

public class Player implements java.io.Serializable {
    // ...
}

public class Location implements java.io.Serializable {
    // ...
}
问题是映射中的对象不是,因此您也必须使它们可序列化

public class SerializedPlayer extends Player implements Serializable {
    public SerializedPlayer() {}
    public SerializedPlayer(Player playerToClone) {
        this.setField1(playerToClone.getField1());
        // Set all the fields
    }
}
添加到地图时:

map.put(new SerializedPlayer(player), new SerializedLocation(location));

当实例需要具有
Serializable
接口时,将引发
notserializableeexception

class YourClass implements Serializable {
    // ...
}

要使
Location
可序列化,我建议使用。 我自己写的,只是要求你在代码中给我信用

package com.github.JamesNorris.Class;

import java.io.Serializable;
import java.util.HashMap;
import java.util.Map;

import org.bukkit.Bukkit;
import org.bukkit.Location;
import org.bukkit.Server;
import org.bukkit.World;

/**
 * The class that allows location to be serialized, and be used
 * in a file. This class should not be changed, otherwise it
 * will not work for older files.
 * 
 * @author Jnorr44
 */

public final class SerializableLocation implements Serializable {
    private static final long serialVersionUID = 8650311534439769069L;

    private final String world;
    private final String uuid;
    private final double x, y, z;
    private final float yaw, pitch;
    private transient Location loc;

    /**
     * Creates a new SerializableLocation instance of any org.bukkit.Location.
     * 
     * @param l
     */

    public SerializableLocation(Location l) {
        this.world = l.getWorld().getName();
        this.uuid = l.getWorld().getUID().toString();
        this.x = l.getX();
        this.y = l.getY();
        this.z = l.getZ();
        this.yaw = l.getYaw();
        this.pitch = l.getPitch();
    }

    /**
     * Gets the org.bukkit.Location back from any SerializableLocation.
     * 
     * @param l
     * @return
     */

    public static Location returnLocation(SerializableLocation l) {
        float pitch = l.pitch;
        float yaw = l.yaw;
        double x = l.x;
        double y = l.y;
        double z = l.z;
        World world = Bukkit.getWorld(l.world);
        Location location = new Location(world, x, y, z, yaw, pitch);
        return location;
    }

    // FROM HERE ON NEEDS DOC NOTES

    public SerializableLocation(Map<String, Object> map) {
        this.world = (String) map.get("world");
        this.uuid = (String) map.get("uuid");
        this.x = (Double) map.get("x");
        this.y = (Double) map.get("y");
        this.z = (Double) map.get("z");
        this.yaw = ((Float) map.get("yaw")).floatValue();
        this.pitch = ((Float) map.get("pitch")).floatValue();
    }

    public final Map<String, Object> serialize() {
        Map<String, Object> map = new HashMap<String, Object>();
        map.put("world", this.world);
        map.put("uuid", this.uuid);
        map.put("x", this.x);
        map.put("y", this.y);
        map.put("z", this.z);
        map.put("yaw", this.yaw);
        map.put("pitch", this.pitch);
        return map;
    }

    public final Location getLocation(Server server) {
        if (loc == null) {
            World world = server.getWorld(this.uuid);
            if (world == null) {
                world = server.getWorld(this.world);
            }
            loc = new Location(world, x, y, z, yaw, pitch);
        }
        return loc;
    }
}

之所以需要这样做,是因为
Location
包含
world
x
y
z
yaw
pitch
,其中
world
是不可序列化的。

事实上,您正在尝试序列化
Player
Location
,尝试实现新类等,但为什么要这样做

public class Player implements java.io.Serializable {
    // ...
}

public class Location implements java.io.Serializable {
    // ...
}
最好不要存储
Player
Location
对象,而是存储它们的字符串表示形式。例如,您可以使用
Player.getName()
和类似
“world:100:65:100”
的内容作为位置(这样我们就可以通过
String.split(“:”
)轻松获取所有数据。我认为这是一种更好的方法