杰克逊@JsonCreator无法识别的财产 [英] Unrecognised Property with Jackson @JsonCreator

问题描述

我正在尝试使用Jackson将一些JSON转换为包含一些简单Strings和另一个类的类的实例，而我正在使用@JsonCreator.看来Jackson不能创建另一个类的实例.

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.

问题是，当我在测试中运行此代码时:

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);

我收到以下异常:

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

我要在简单测试中解析的JSON看起来像这样:

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

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

我有一个看起来像这样的玩家"类

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
}

还有另一个看起来像这样的字符"类

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
}

对于像这样的少量数据，将有更好的方法来构建整个结构，但是我认为这只是出于示例的目的.我拥有的实际代码更加复杂，但我想举一个简单的例子.

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.

我认为我已经弄乱了Jackson的注解，但是我不确定自己做错了什么.

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

推荐答案

您需要在Player上指定与您的JSON输入匹配的创建者.例如:

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;
}

旁注，您可以像 this 那样构造枚举，而Jackson将为您完成从字符串到枚举的转换

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

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