CODEWITHSUNDEEP
javajsonjackson
Héctor
Héctor

Reputation: 26054

Is it possible to group JSON fields into one property using Jackson?

I'm wondering if there is any way to deserialize several JSON fields to just one Java property. E.g. given this JSON:

{
  "id" : "1",
  "name" : "Bartolo",
  "address" : "whatever",
  "phone" : "787312212"
}

deserialize it to this class:

public class Person {

    public String id;
    public String name:

    @JsonProperty(names = {"address", "phone"}) //something like this
    public String moreInfo;

}

so moreInfo equals to "whatever, 787312212" or something similar.

Is this possible without using custom deserializer?

Upvotes: 0

Views: 1130

Answers (2)

d_ser
d_ser

Reputation: 69

Another solution, if you don't want to know/handle other fields in the object, but decided to still receive these fields (maybe for logging purposes), then you can put them in a key-value store(Map)

@Getter
private final Map<String, Object> otherFields = new HashMap<>();

@JsonAnySetter
public void set(String name, Object value) {
    otherFields.put(name, value);
}

Note that if you have any field with the same name as the Map field(like 'otherFields' in the example above), then you can get MismatchedInputException

Upvotes: 0

NickAth
NickAth

Reputation: 1112

You could use the @JsonCreator annotation like following:

String json = {"id" : "1",  "name" : "Bartolo", "address" : "whatever", "phone" : "787312212" }

ObjectMapper mapper = new ObjectMapper();
Person person = mapper.readValue(json , Person.class);

and in the constructor of your Person class add this

@JsonCreator
public Person(@JsonProperty("address") String address, @JsonProperty("phone") String phone) {
    this.moreInfo = address + "," phone;
}

Upvotes: 1

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