Unrecognised Property with Jackson @JsonCreator

Question

I'm trying to use Jackson to convert some JSON into an instance of a class that contains some simple Strings and another class, which I'm using @JsonCreator for. It seems that Jackson isn't able to create the instance of the other class.

The problem is that when I run this code as part of a test:

ObjectMapper mapper = new ObjectMapper();
Player player = mapper.readValue(json.toString(), Player.class);

I get the following exception:

com.fasterxml.jackson.databind.exc.UnrecognizedPropertyException: Unrecognized field "characterClass" (class xxx.xxx.Player), not marked as ignorable (2 known properties: "name", "character"])

The JSON I'm trying to parse in my simple test looks like this:

{
    "name": "joe",
    "characterClass": "warrior",
    "difficulty": "easy",
    "timesDied": 2
}

I have a class 'Player' that looks a bit like this

public class Player {

    @JsonProperty("name")
    private String playerName;

    @JsonProperty // <-- This is probably wrong
    private Character character;

    // Some getters and setters for those two fields and more
}

And another class 'Character' that looks like this

public class Character{
    private PlayerClass playerClass;
    private Difficulty difficulty;
    private int timesDied;

    @JsonCreator
    public Character(@JsonProperty("characterClass") String playerClass,
                       @JsonProperty("difficulty") String diff,
                       @JsonProperty("timesDied") int died) {

        // Validation and conversion to enums

        this.playerClass = PlayerClass.WARRIOR;
        this.difficulty = Difficulty.EASY;
        this.timesDied = died;
    }

    // Again, lots of getters, setters, and other stuff
}

For small sets of data like this there would be better ways to structure the whole thing, but I think this works for the purposes of an example. The real code I have is more complex but I wanted to make simple example.

I think I've messed up the Jackson annotations, but I'm not sure what I've done wrong.

score 1 · Accepted Answer · edited May 23 '17 at 11:44

1

You need to specify a creator on Player that matches your JSON input. For example:

@JsonCreator
public static Player fromStringValues(@JsonProperty("name") String name,
                                      @JsonProperty("characterClass") String characterClass,
                                      @JsonProperty("difficulty") String difficulty,
                                      @JsonProperty("timesDied") Integer timesDied) {
    Player player = new Player();
    player.setPlayerName(name);
    player.setCharacter(new Character(characterClass, difficulty, timesDied));
    return player;
}

A side note, you can structure your enums like this and Jackson will do the conversion from string to enum for you.

edited May 23 '17 at 11:44

Community

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answered Jul 09 '15 at 18:27

Sam Berry

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Is there a way to do this without having to list every property as an argument? I was hoping for something that just creates the 'Character' class based on the three properties it needs and nothing else. – Charlie H Jul 10 '15 at 09:54
@CharlieHurrell That would be convenient, but no you have to provide Jackson with a creator if you are not following default deserialization. For example, how would it know what to do if you had a `name` field on `Player` and `Character`. Instructions must be provided. – Sam Berry Jul 10 '15 at 14:24

Unrecognised Property with Jackson @JsonCreator

1 Answers1