How to deserialize a class with overloaded constructors using JsonCreator

Answer 1

If I got right what you are trying to achieve, you can solve it without a constructor overload.

If you just want to put null values in the attributes not present in a JSON or a Map you can do the following:

@JsonIgnoreProperties(ignoreUnknown = true)
public class Person {
    private String name;
    private Integer age;
    public static final Integer DEFAULT_AGE = 30;

    @JsonCreator
    public Person(
        @JsonProperty("name") String name,
        @JsonProperty("age") Integer age) 
        throws IllegalArgumentException {
        if(name == null)
            throw new IllegalArgumentException("Parameter name was not informed.");
        this.age = age == null ? DEFAULT_AGE : age;
        this.name = name;
    }
}

That was my case when I found your question. It took me some time to figure out how to solve it, maybe that's what you were tring to do. @Brice solution did not work for me.