Jackson vector serialization exception - serialization

Question

I have the following code with a simple class and a method for writing and then reading:
ObjectMapper mapper = new ObjectMapper();
try{
DataStore testOut = new DataStore();
DataStore.Checklist ch1 = testOut.addChecklist();
ch1.SetTitle("Checklist1");
String output = mapper.writeValueAsString(testOut);
JsonNode rootNode = mapper.readValue(output, JsonNode.class);
Map<String,Object> userData = mapper.readValue(output, Map.class);
}
public class DataStore {
public static class Checklist
{
public Checklist()
{
}
private String _title;
public String GetTitle()
{
return _title;
}
public void SetTitle(String title)
{
_title = title;
}
}
//Checklists
private Vector<Checklist> _checklists = new Vector<Checklist>();
public Checklist addChecklist()
{
Checklist ch = new Checklist();
ch.SetTitle("New Checklist");
_checklists.add(ch);
return ch;
}
public Vector<Checklist> getChecklists()
{
return _checklists;
}
public void setChecklists(Vector<Checklist> checklists)
{
_checklists = checklists;
}
}
The line:
String output = mapper.writeValueAsString(testOut);
causes an exception that has had me baffled for hours and about to abandon using this at all.
Any hints are appreciated.
Here is the exception:
No serializer found for class DataStore$Checklist and no properties discovered to create BeanSerializer (to avoid exception, disable SerializationConfig.Feature.FAIL_ON_EMPTY_BEANS) ) (through reference chain: DataStore["checklists"]->java.util.Vector[0])

Answer 1

There are multiple ways to do it, but I will start with what you are doing wrong: your naming of getter and setter method is wrong -- in Java one uses "camel-case", so you should be using "getTitle". Because of this, properties are not found.
Besides renaming methods to use Java-style names, there are alternatives:
You can use annotation JsonProperty("title") for GetTitle(), so that property is recognized
If you don't want the wrapper object, you could alternatively just add #JsonValue for GetTitle(), in which case value used for the whole object would be return value of that method.

Answer 2

The answer seems to be: You can't do that with Json. I've seen comments in the Gson tutorial as well, that state that some serialization just doesn't work. I downloaded XStream and spat it out with XML in a few minutes of work and a lot less construction around what I really wanted to persist. In the process, I was able to delete a lot of code.