Using Jackson to (De)-serialize a Scala Case Class

Question

I tested out the serialization of a Scala case class using Jackson.

DeserializeTest.java

    public static void main(String[] args) throws Exception { // being lazy to catch-all

        final ObjectMapper mapper          = new ObjectMapper();
        final ByteArrayOutputStream stream = new ByteArrayOutputStream();

        mapper.writeValue(stream, p.Foo.personInstance());

        System.out.println("result:" +  stream.toString());
    }
}

Foo.scala

object Foo {
  case class Person(name: String, age: Int, hobbies: Option[String])
  val personInstance = Person("foo", 555, Some("things"))
  val PERSON_JSON = """ { "name": "Foo", "age": 555 } """
}

When I ran the above main of the Java class, an exception was thrown:

[error] Exception in thread "main" org.codehaus.jackson.map.JsonMappingException: 
 No serializer found for class p.Foo$Person and no properties discovered 
 to create BeanSerializer (to avoid exception, 
 disable SerializationConfig.Feature.FAIL_ON_EMPTY_BEANS) )

How can I (de)-serialize Scala case classes?

Answer 1

Jackson is expecting your class to be a JavaBean, which means its expects the class to have a getX() and/or setX() for every property.

Option 1

You can create JavaBean classes in Scala using the annotation BeanProperty.

Example

case class Person(
   @BeanProperty val name: String, 
   @BeanProperty val age: Int, 
   @BeanProperty val hobbies: Option[String]
)

In this case a val will mean only a getter is defined. If you want setters for deserialization you defined the properties as var.

Option 2

While option 1 will work, if you really want to use Jackson there are wrappers that allow it to deal with Scala classes like FasterXML's scala module which might be a better approach. I haven't used it as I've just been using the Json library built in to play.