How to de/serialize an immutable object without default constructor using ObjectMapper?

Question

I want to serialize and deserialize an immutable object using com.fasterxml.jackson.databind.ObjectMapper.

The immutable class looks like this (just 3 internal attributes, getters and constructors):

public final class ImportResultItemImpl implements ImportResultItem {

    private final ImportResultItemType resultType;

    private final String message;

    private final String name;

    public ImportResultItemImpl(String name, ImportResultItemType resultType, String message) {
        super();
        this.resultType = resultType;
        this.message = message;
        this.name = name;
    }

    public ImportResultItemImpl(String name, ImportResultItemType resultType) {
        super();
        this.resultType = resultType;
        this.name = name;
        this.message = null;
    }

    @Override
    public ImportResultItemType getResultType() {
        return this.resultType;
    }

    @Override
    public String getMessage() {
        return this.message;
    }

    @Override
    public String getName() {
        return this.name;
    }
}

However when I run this unit test:

@Test
public void testObjectMapper() throws Exception {
    ImportResultItemImpl originalItem = new ImportResultItemImpl("Name1", ImportResultItemType.SUCCESS);
    String serialized = new ObjectMapper().writeValueAsString((ImportResultItemImpl) originalItem);
    System.out.println("serialized: " + serialized);

    //this line will throw exception
    ImportResultItemImpl deserialized = new ObjectMapper().readValue(serialized, ImportResultItemImpl.class);
}

I get this exception:

com.fasterxml.jackson.databind.JsonMappingException: No suitable constructor found for type [simple type, class eu.ibacz.pdkd.core.service.importcommon.ImportResultItemImpl]: can not instantiate from JSON object (missing default constructor or creator, or perhaps need to add/enable type information?)
 at [Source: {"resultType":"SUCCESS","message":null,"name":"Name1"}; line: 1, column: 2]
    at 
... nothing interesting here

This exception asks me to create a default constructor, but this is an immutable object, so I don't want to have it. How would it set the internal attributes? It would totally confuse the user of the API.

So my question is: Can I somehow de/serialize immutable objects without default constructor?

While desrializing, the desrializer doesn't know about any of your constructor, so it hits default constructor. Due to this you have to create a default constructor, that won't change the immutability. Also I see the class is not final, why so? anybody can override the functionality isn't it? — Abhinaba Basu, Jun 01 '15 at 08:04
Class is not final, because I forgot to write it there, thank you for noticing :) — Michal Krasny, Jun 01 '15 at 08:15

score 191 · Accepted Answer · answered Jun 01 '15 at 08:12

191

To let Jackson know how to create an object for deserialization, use the @JsonCreator and @JsonProperty annotations for your constructors, like this:

@JsonCreator
public ImportResultItemImpl(@JsonProperty("name") String name, 
        @JsonProperty("resultType") ImportResultItemType resultType, 
        @JsonProperty("message") String message) {
    super();
    this.resultType = resultType;
    this.message = message;
    this.name = name;
}

answered Jun 01 '15 at 08:12

Forketyfork

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+1 thanks Sergey. It worked.. however I was having problem even after using JsonCreator annotation.. but after some review I realized that I forgot to use RequestBody annotation in my resource. Now it works.. Thanx once again. – Rahul Mar 14 '16 at 14:07
8

What if you can't add a default constructor, or the `@JsonCreator` and `@JsonProperty` annotations because you don't have access to the POJO to change it? Is there still a way to deserialize the object? – Jack Straw Aug 29 '17 at 02:01
1

@JackStraw 1) subclass from the object and override all getters and setters. 2) write your own serializer/deserializer for the class and use it (search for "custom serializer" 3) wrap the object and write a serializer/deserializer just for the one property (this may let you do less work than writing a serializer for the class itself, but maybe not). – ErikE Sep 20 '17 at 02:11
+1 your comment saved me! Every solution so far that i was finding was create the default constructor... – Nilson Aguiar Feb 16 '20 at 16:50
1

How can we do this if we are using @Builder Annotation from Lombok? – manikanta nvsr Dec 23 '20 at 06:57
@manikantanvsr you can add a constructor and move the `@Builder` annotation to this constructor, instead of the class. – Forketyfork Dec 24 '20 at 07:29
1

@Sergei There is one annotation "Jacksonized". We can use it on top of the builder annotation. – manikanta nvsr Dec 24 '20 at 07:33
Great answer. Followup question, what if one of the params I get in the ctor should be set to an inner object? Something like: `@JsonCreator MyClass(String innerField) { this.someObject.setInnerField(innerField);}` Can we use something nested like `JsonProperty("someObject.innerField")` ? – Nom1fan Aug 11 '22 at 07:51

tkruse · Answer 2 · 2017-10-17T05:36:20.120

53

You can use a private default constructor, Jackson will then fill the fields via reflection even if they are private final.

EDIT: And use a protected/package-protected default constructor for parent classes if you have inheritance.

edited Oct 17 '17 at 05:36

answered Oct 06 '17 at 08:29

tkruse

10,222
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Just to build on this answer, if the class you want to deserialise extends another class then you can make the default constructor of the super class (or both classes) protected. – KWILLIAMS Oct 13 '17 at 14:10

score 29 · Answer 3 · answered Apr 27 '20 at 21:52

The first answer of Sergei Petunin is right. However, we could simplify code with removing redundant @JsonProperty annotations on each parameter of constructor.

It can be done with adding com.fasterxml.jackson.module.paramnames.ParameterNamesModule into ObjectMapper:

new ObjectMapper()
        .registerModule(new ParameterNamesModule(JsonCreator.Mode.PROPERTIES))

(Btw: this module is registered by default in SpringBoot. If you use ObjectMapper bean from JacksonObjectMapperConfiguration or if you create your own ObjectMapper with bean Jackson2ObjectMapperBuilder then you can skip manual registration of the module)

For example:

public class FieldValidationError {
  private final String field;
  private final String error;

  @JsonCreator
  public FieldValidationError(String field,
                              String error) {
    this.field = field;
    this.error = error;
  }

  public String getField() {
    return field;
  }

  public String getError() {
    return error;
  }
}

and ObjectMapper deserializes this json without any errors:

{
  "field": "email",
  "error": "some text"
}

How do you use this, if you dont manually initialize ObjectMapper? — joachim, May 10 '21 at 15:56
I inject it with Spring Context. Spring boot provides a bean of ObjectMapper — Vladyslav Diachenko, May 11 '21 at 16:55

score 5 · Answer 4 · answered Sep 08 '21 at 13:15

It's 2021, I had the same issue. Unfortunately, the previous answers in this thread weren't helpful in my case, because:

I need to deserialize java.net.HttpCookie class object. I can't modify it.
Due to some reasons, I use jackson-databind-2.11.0 version. ParameterNamesModule has been added in 3+ version.

So, here's the solution I've found for my case. You can just use DeserializationProblemHandler:

    objectMapper.addHandler(new DeserializationProblemHandler() {


        @Override
        public Object handleMissingInstantiator(DeserializationContext ctxt, Class<?> instClass, ValueInstantiator valueInsta, JsonParser p, String msg) throws IOException {
            return super.handleMissingInstantiator(ctxt, instClass, valueInsta, p, msg);
        }

    });

Just return the object which you're expecting

Excellent Answer, Works well – Pavlo Plynko Sep 09 '21 at 06:48 — Pavlo Plynko, Sep 09 '21 at 06:48

score 1 · Answer 5 · answered Jun 28 '22 at 13:00

Since Java 14 and Jackson 2.12.0 you can use a record class like this:

public record ImportResultItemImpl(String name, ImportResultItemType resultType, String message) implements ImportResultItem {
    
    public ImportResultItemImpl(String name, ImportResultItemType resultType) {
        // calling the default constructor
        this(name, resultType, null);
    }
}

Also, you will have to refactor your interface and usage because record's getters do not start with get or is:

public interface ImportResultItem {

    String name();

    ImportResultItemType resultType();

    String message();
}

How to de/serialize an immutable object without default constructor using ObjectMapper?

5 Answers5

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