如何将 map 几个 json 字段转换为单个 java Z1D78DC8ED51214E518B5114AE24490？

Question

How to map this JSON to this Java object where the habits Map will have the keys HabitName.FITNESS and HabitName.MATHEMATICS with the corresponding values? Constructor, getters and setters are omitted to readability.

{
  "habits": {
    "fitness": {
      "duration": "1 week",
      "score": "2.1"
    },
    "mathematics": {
      "duration": "2 weeks",
      "score": "2.4"
    }
  }
}

public class IncomingJson {
  Map<HabitName, HabitDesc> habits;

  public enum HabitName {
    @JsonProperty("fitness") FITNESS,
    @JsonProperty("mathematics") MATHEMATICS
  }
  
  public static class HabitDesc {
    String duration;
    String score;
  }
}

At runtime I get the following Exception:

com.fasterxml.jackson.databind.exc.UnrecognizedPropertyException: Unrecognized field "fitness" (class IncomingJson), not marked as ignorable (one known property: "habits") 

Answer 1

The solution was close. You can deserialize this JSON into the following Java class easily.

{
    "root" : {
        "habits": {
            "fitness": {
                "duration": "1 week",
                "score": "2.1"
            },
            "mathematics": {
                "duration": "2 weeks",
                "score": "2.4"
            }
        }
    }
}

@Getter
@RequiredArgsConstructor
@JsonRootName("root")
publi class Habits {
    private final Map<HabitName, HabitDesc> habits = new HashMap<>();
    public enum HabitName {
        @JsonProperty("fitness") FITNESS,
        @JsonProperty("mathematics") MATHEMATICS
    }
    @Getter
    @RequiredArgsConstructor
    public static class HabitDesc {
        @JsonProperty("duration") private final String duration;
        @JsonProperty("score") private final String score;
    }
}

IMPORTANT : The above class relies on lombok.anyconstructor.addconstructorproperties=true in the lombok.config file in the same package as Habits.class .

So your deserialization code will look like this

var objectMapper = new ObjectMapper();
objectMapper.configure(DeserializationFeature.UNWRAP_ROOT_VALUE, true);
Habits = objectMapper.readValue(JSON_STRING, Habits.class);