Question 341940 in Json Question posted in
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Add new Object to the JSON output

I have a POJO like this:

public class Person {
  public long id;
  public String someOtherId;
  public String vollName;
  public int alter;
}

I am converting the POJO to JSON using the mixin:

@JsonIncludeProperties( { "id", "name", "age", "customFields" } )
abstract class PersonM extends Person {
  @JsonProperty( "id" ) String someOtherId;
  @JsonProperty( "name" ) String vollName;
  @JsonProperty( "age" ) String alter;
}

...

ObjectMapper mapper = new ObjectMapper();
mapper.addMixIn( Person.class, PersonM.class );
String body = mapper.writeValueAsString( person );

The code works like charm and produces a JSON of:

{
  "id": "xxxxxx",
  "name": "Aaaa Bbbb",
  "age": 111
}

Now I need to extend the JSON to contain a customFields element:

{
  "id": "xxxxxx",
  "name": "Aaaa Bbbb",
  "age": 111,
  "customFields": {
    "originalId": 333333
  }
}

so that the content of that new element is based on other fields of the same POJO, and would look like so as a getter:

public class Person {
 ...
 public Map getCustomFields() { return Map.of( "originalId", id ); }
}

I don’t want to include this or similar methods into my POJO for a number of reasons and would like to do it on StdSerializer<>-level.

On the Internet I could not find any example on adding another JSON element to the existing fields, all samples show building JSON from scratch instead.

Hence the question: How to add a JSON Object to existing output without "re-inventing the wheel" over and over again?

Tags:

2

Answers


  1. 0

    create a custom PersonDeserializer by extending JsonDeserializer. Inside the deserialize method, we extract the id, name, and age values from the JSON node. Then, we create a Person object and set its properties accordingly. Additionally, we create a customFields map and populate it with the desired custom field values. Finally, we set the customFields map on the Person object.

    import com.fasterxml.jackson.core.JsonParser;
import com.fasterxml.jackson.databind.DeserializationContext;
import com.fasterxml.jackson.databind.JsonDeserializer;
import com.fasterxml.jackson.databind.JsonNode;
import com.fasterxml.jackson.databind.node.ObjectNode;

import java.io.IOException;
import java.util.HashMap;
import java.util.Map;

public class PersonDeserializer extends JsonDeserializer<Person> {

    @Override
    public Person deserialize(JsonParser jsonParser, DeserializationContext deserializationContext) throws IOException {
        ObjectNode node = jsonParser.getCodec().readTree(jsonParser);
        long id = node.get("id").asLong();
        String name = node.get("name").asText();
        int age = node.get("age").asInt();

        Person person = new Person();
        person.setId(id);
        person.setName(name);
        person.setAge(age);

        Map<String, Object> customFields = new HashMap<>();
        customFields.put("originalId", id);
        person.setCustomFields(customFields);

        return person;
    }
}

    To use this custom deserializer, you need to register it with the ObjectMapper:

    ObjectMapper mapper = new ObjectMapper();
SimpleModule module = new SimpleModule();
module.addDeserializer(Person.class, new PersonDeserializer());
mapper.registerModule(module);

String json = mapper.writeValueAsString(person);

    or try this :

      class Person {
            public long id;
            public String someOtherId;
            public String vollName;
            public int alter;
        }

        // Create an instance of the original POJO
        Person person = new Person();
        person.id = 123;
        person.someOtherId = "abc";
        person.vollName = "John Doe";
        person.alter = 30;

        // Create a custom map for additional properties
        Map<String, Object> additionalProperties = new HashMap<>();
        additionalProperties.put("originalId", 333333);

        // Create an instance of the ObjectMapper
        ObjectMapper mapper = new ObjectMapper();

        // Convert the POJO to JSON
        String json = mapper.writeValueAsString(person);

        // Merge the additional properties with the original JSON
        String mergedJson = mergeJson(json, additionalProperties);
        System.out.println(mergedJson);
    }

    private static String mergeJson(String originalJson, Map<String, Object> additionalProperties) throws Exception {
        ObjectMapper mapper = new ObjectMapper();
        Map<String, Object> mergedProperties = mapper.readValue(originalJson, Map.class);

        // Merge the additional properties with the original properties
        mergedProperties.putAll(additionalProperties);
        return mapper.writeValueAsString(mergedProperties);
    }
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  2. 0

    You can add elements to your json, your only need convert from String, did you try this?

    ObjectNode json=(ObjectNode) mapper.readTree(body);
json.set("CustomFields", mapper.valueToTree(Map.of( "originalId", person.getId())));
body=json.toString();
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