杰克逊JSON：将映射键作为包含对象的属性

Question

给定这样的结构：

{
  "nameOfObject": { "score": 100 },
  "anotherObject": { "score": 30 }
}

能否将此映射为：

class Container {
  Map<String, ScoreKeeper> scoreKeepers;
}

class ScoreKeeper {
  String name;
  int score;
}

你希望让ScoreKeeper实例的name属性分别设置为"nameOfObject"和"anotherObject"，是吗？

- Nick Spacek

---

你可能需要创建一个中间对象进行序列化，然后将其转换为你想要的表示形式。 - Bart

Answer 1

我坚信将POJO（普通Java对象）与外部化分离。将JSON读入Map中，然后像这样构建Container/ScoreKeeper对象（抱歉拼写错误）：

mapper = new ObjectMapper();

Map<String,Object> data = mapper.readValue(inputstream, Map.class);

Container c = new Container();

for(Map.Entry<String, Object> me : data.entrySet()) {
    String key = me.getKey();
    Map info = (Map) me.getValue();

    ScoreKeeper sk = new ScoreKeeper();
    sk.setName(key);
    Integer q = info.get("score");
    sk.setScore(q);

    c.put(key, sk);
}

- Buzz Moschetti

---

当然。我实际上只是在寻找一种简洁的方法来反序列化现有的API。不过还是谢谢你的建议。 - Nick Spacek

Answer 2

另一种解决方案是使用自定义反序列化器将键name设置在值对象上：

@Test
public void test() throws JsonParseException, JsonMappingException, IOException {
    ObjectMapper mapper = new ObjectMapper();

    Data data = mapper.readValue("{\"users\": {\"John\": {\"id\": 20}, \"Pete\": {\"id\": 30}}}", Data.class);

    assertEquals(20, data.users.get("John").id);
    assertEquals(30, data.users.get("Pete").id);
    assertEquals("John", data.users.get("John").name);
    assertEquals("Pete", data.users.get("Pete").name);
}

public static class Data {
    @JsonDeserialize(contentUsing = Deser.class)
    public Map<String, User> users;
}

public static class User {
    public String name;
    public int id;
}

public static class Deser extends JsonDeserializer<User> {

    @Override
    public User deserialize(JsonParser p, DeserializationContext ctxt) throws IOException, JsonProcessingException {
        String name = ctxt.getParser().getCurrentName();

        User user = p.readValueAs(User.class);

        user.name = name;  // This copies the key name to the value object

        return user;
    }
}

- john16384

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我无法使用原始问题中请求的JSON使这种方法起作用，只有在包含“nameOfObject”和“anotherObject”的根节点存在时，此方法才有效。 - Nick

Answer 3

这是@Buzz Moschetti的改进版本，它使用Jackson的ObjectMapper.convertValue()来处理属性解析。

ObjectMapper mapper = new ObjectMapper();

Map<String,Object> data = mapper.readValue(inputstream, Map.class);

Container c = new Container();

for(Map.Entry<String, Object> entry : data.entrySet()) {
    String name = entry.getKey();
    ScoreKeeper sk = mapper.convertValue(entry.getValue(), ScoreKeeper.class);
    sk.name = name;
    c.scoreKeepers.put(name, sk);
}

- Nick