Serialization - not working

Question

I create class with methods like a How to serialize an object into a string and it every say error "java.lang.ClassCastException: java.lang.String cannot be cast to Myclass"

My codes: 1)

import java.io.ByteArrayInputStream;
import java.io.ByteArrayOutputStream;
import java.io.IOException;
import java.io.ObjectInputStream;
import java.io.ObjectOutputStream;
import java.io.Serializable;

import javax.xml.bind.DatatypeConverter;

public class Serialization {

    public static Object fromString(String s) throws IOException,
            ClassNotFoundException {
        byte[] data = DatatypeConverter.parseBase64Binary(s);
        ObjectInputStream ois = new ObjectInputStream(new ByteArrayInputStream(
                data));
        Object o = ois.readObject();
        ois.close();
        return o;
    }

    public static String toString(Serializable o) throws IOException {
        ByteArrayOutputStream baos = new ByteArrayOutputStream();
        ObjectOutputStream oos = new ObjectOutputStream(baos);
        oos.writeObject(o);
        oos.close();
        return DatatypeConverter.printBase64Binary(baos.toByteArray());
    }
}

2) - calling

MyClass hl = (MyClass) Serialization.fromString(items
        .getString("data"));
hl.load(); // this is my method from class

3) MyClass - Hologram

public class Hologram implements Serializable {

 /**
 * 
 */
private static final long serialVersionUID = 1L;

private Location loc;
private String name;
private String displayname;
public ArmorStand stand;

public Hologram(String name, String displayname, Location loc) {
this.loc = loc;
this.name = name;
this.displayname = displayname;

ArmorStand as = (ArmorStand) loc.getWorld().spawnEntity(loc,
    EntityType.ARMOR_STAND);

as.setGravity(false);
as.setCanPickupItems(false);
as.setCustomName(displayname);
as.setCustomNameVisible(true);
as.setVisible(false);

this.stand = as;

HologramManager.holograms.put(name, this);

}

public void move(Location loc) {
this.loc = loc;

stand.teleport(loc);
}

public Location getLocation() {
return this.loc;
}

public void remove() {
stand.remove();
HologramManager.holograms.remove(name);
}

public void removeHologram() {
HologramManager.remove(name);
}

public void changeName(String name) {
HologramManager.holograms.remove(this.name);
this.name = name;
HologramManager.holograms.put(name, this);
}

public void changeDisplayName(String displayName) {
this.displayname = displayName;
stand.setCustomName(displayname);
stand.setCustomNameVisible(true);
}

public void load() {
//todo
}

}

Please post a short but *complete* program demonstrating the problem. We can't see how you came up with the string to start with. (I'd also strongly recommend avoiding Java binary serialization, but that's a different matter.) — Jon Skeet, Jun 11 '15 at 12:29
If you want to deserialize like that, you'll need to read the bytes back into a `java.io.ObjectInputStream` and `readObject()`. — javabrett, Jun 11 '15 at 12:31

score 1 · Accepted Answer · answered Jun 11 '15 at 12:48

1

Based on the linked answer, the problem most likely lies in the code you aren't showing us. When you serialize your MyClass object, you are probably doing something like this:

MyClass hl;
String base64String = Serialization.toString(hl.toString());

However you should be calling it like this:

MyClass hl;
String base64String = Serialization.toString(hl);

If you pass a String to the serialization function, you'll get a String back when you call Serialization.fromString(). You want to get an object back that you can cast to a MyClass instance, so pass one of those into Serialization.toString().

answered Jun 11 '15 at 12:48

samgak

23,944
4
60
82

thanks you, now it say java.io.NotSerializableException: org.bukkit.Location - what do now? – DenOwq DO Jun 11 '15 at 12:51
1

Make the Location class Implement Serializable class. – Kalyan Chavali Jun 11 '15 at 12:56
In addition to your making your Hologram class implement Serializable, all the member variables need to implement it too. You can either subclass them and implement Serializable, or declare them as transient if they don't need to be serialized – samgak Jun 11 '15 at 12:58
A location is existing class in another Lib(spigot[bukkit]) – DenOwq DO Jun 11 '15 at 12:58
Can you subclass it and implement Serializable in the subclass? – samgak Jun 11 '15 at 13:01
Location is interface.. :/ – DenOwq DO Jun 11 '15 at 13:13
I dont have any time to doing this... its closed – DenOwq DO Jun 11 '15 at 13:16

score 0 · Answer 2 · answered Jun 11 '15 at 12:41

0

The fromString() method in Serilization returns an Object, which you wouldnt be able to cast to the class MyClass. The below line is causing the classCastException

MyClass hl = (MyClass) Serialization.fromString(items
    .getString("data"));

answered Jun 11 '15 at 12:41

Kalyan Chavali

1,330
8
24

And what do you propose for solution ? – JFPicard Jun 11 '15 at 12:42
I dont know how to repair that – DenOwq DO Jun 11 '15 at 12:47
Read an object , Object obj = Serialization.fromString(items .getString("data"));. Its upto you on what you decide to do with it ;) – Kalyan Chavali Jun 11 '15 at 12:52

Serialization - not working

2 Answers2