无法识别属性与 Jackson @JsonCreator

Question

我正在尝试使用 Jackson 将一些 JSON 转换为一个 class 的实例，其中包含一些简单的字符串和另一个 class，我正在使用 @JsonCreator为了。杰克逊似乎无法创建另一个 class.

的实例

问题是，当我运行这段代码作为测试的一部分时：

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

我得到以下异常：

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

我在简单测试中尝试解析的 JSON 如下所示：

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

我有一个 class 'Player' 看起来有点像这样

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
}

还有一个 class 'Character' 看起来像这样

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
}

对于像这样的小数据集，会有更好的方法来构建整个数据，但我认为这只是为了举例。我的实际代码比较复杂，但我想做一个简单的例子。

我想我弄乱了 Jackson 注释，但我不确定我做错了什么。

Answer 1

您需要在 Player 上指定与您的 JSON 输入匹配的创作者。例如：

@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 那样构造您的枚举，Jackson 会为您完成从字符串到枚举的转换。

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