JSON serialization - how not to serialize one class property with the whole structure

Question

I want to serialize some string json (I use Spring Boot):

{
  "commandId": "34d3a914-a3d7-112a-bs37-3452ac130002",
  "status": "SUCCESS",
  "response": {
    "prop1": "",
    "prop2": "",
    "prop3": "true",
    "prop4": ""
  }
}

My mapper:

private static final ObjectMapper mapper = new ObjectMapper();

public static <T> T fromJson(String json, TypeReference<T> type) {
    T t;
    try {
      t = mapper.reader().forType(type).readValue(json);
    } catch (IOException e) {
      String message = "Error converting from json";
      log.error(message, e);
      throw new IllegalArgumentException(message);
    }

The usage of the mapper:

final Command command =
              JsonUtils.fromJson(json, new TypeReference<Command>() {});

Command class:

public class Command {
  private UUID commandId;
  private status status;
  private String response;
  private String error;
}

Currently I get the error:

com.fasterxml.jackson.databind.exc.MismatchedInputException: Cannot deserialize instance of `java.lang.String` out of START_OBJECT token

I want to save in Mongo db response field as string and later on after receiving that object from DB I can deserialize the whole object. Is there any annotation to tell the Json mapper to not to serialize the structure inside the response field?

score 0 · Answer 1 · answered Jul 05 '20 at 12:35

You can write a custom deserializer for response field to deserialize reponse node object as string. And use it one reponnse field using @JsonDeserialize.

public class Command {
  ...
  @JsonDeserialize(using = RawJsonDeserializer.class)
  private String response;
  ...
}

public class RawJsonDeserializer extends JsonDeserializer<String> {
    @Override
    public String deserialize(JsonParser jp, DeserializationContext ctxt)
           throws IOException, JsonProcessingException {
        ObjectMapper mapper = (ObjectMapper) jp.getCodec();
        JsonNode node = mapper.readTree(jp);
        return mapper.writeValueAsString(node);
    }
}

score 0 · Answer 2 · answered Jul 05 '20 at 12:42

Other than having a deserializer, another simpler approach is using @JsonProperty annotation in your Command class.

public class Command {
  private UUID commandId;
  private String status;
  private String response;
  private String error;

  public UUID getCommandId() {
    return commandId;
  }

  @JsonProperty("commandId")
  public void setCommandId(String commandId) {
    this.commandId = UUID.fromString(commandId);
  }

  public String getStatus() {
    return status;
  }

  public void setStatus(String status) {
    this.status = status;
  }

  public String getResponse() {
    return response;
  }

  @JsonProperty("response")
  private void setResponse(Map<String, String> response) {
    this.response = response.toString();
  }

  public String getError() {
    return error;
  }

  public void setError(String error) {
    this.error = error;
  }
}

Actually now `response` not deserialize as `Map` and you are make toString of map. So it's not json anymore and I think OP want to deserialize again. Then it will not work. — Eklavya, Jul 05 '20 at 14:13

JSON serialization - how not to serialize one class property with the whole structure

2 Answers2