jongo / jackson deserialize scala.option in java

Question

FOUND SOLUTION TO THE PROBLEM For the people who will be stuck like me!: in order to handle third party java or scala Objects for jackson deserialization, you can either use Mixins( but you need to reconfigure the jackson mapper or user Modules) OR you can simply create a class called MyClassDeserializer that extends JsonDeserializer and use the @JsonDeserialize(using = MyClassDeserializer.class) annotation.

exemple :

it's really simple and works like a charm! :)

public class User implements Identity{
       @JsonProperty("_id")
        private String id;
        @JsonDeserialize(using = OptionDeserializer.class)
        public Option<String> email;
    }



 public class OptionDeserializer extends JsonDeserializer<Option> {
        @Override
        public Option deserialize(JsonParser jsonParser, DeserializationContext deserializationContext) throws IOException, JsonProcessingException {
            //code can be improved
ObjectCodec oc = jsonParser.getCodec();
            JsonNode node = oc.readTree(jsonParser);
            return  Option.apply(node.get("email").getText());
        }
    }

hello guys i'm having some bad time trying to deserialize this with jackson using Jongo

public class User implements Identity{
   @JsonProperty("_id")
    private String id;
    public Option<String> email;
}

Option is an abstract type, I'm really new to this, is there a way to actually tell jackson how to translate it?

this User object is correctly saved to mongo, but cannot be read :/ i've spent so many hours trying to understand you guys are my last resort! thanks (keeps looking)

Caused by: com.fasterxml.jackson.databind.JsonMappingException: Can not construct instance of scala.Option, problem: abstract types either need to be mapped to concrete types, have custom deserializer, or be instantiated with additional type information at [Source: de.undercouch.bson4jackson.io.LittleEndianInputStream@6f255853; pos: 237] (through reference chain: models.User["email"])

What is the class `Option` and does it contain default constructor? — Michał Ziober, Nov 10 '13 at 01:31
hello Michal Option class is a scala class :/ i'm using a third parti liobrary that uses scala in my java project, and I need to serialize an object containing an Option field. — popo joe, Nov 10 '13 at 02:07
thx michal i've found the solution to the problem and posted it on my first post for futur users — popo joe, Nov 10 '13 at 03:09
@user2207996 you should post your answer as an answer below, and accept it, so that the question does not show up as an unanswered question in the lists. — Robin Green, Nov 10 '13 at 08:25
No problem. You can post your solution as an answer. It will be the clearest way. — Michał Ziober, Nov 10 '13 at 23:51

Andrew Norman · Answer 1 · 2016-02-19T20:33:02.580

posting popo joe's answer as an answer:

FOUND SOLUTION TO THE PROBLEM For the people who will be stuck like me!: in order to handle third party java or scala Objects for jackson deserialization, you can either use Mixins( but you need to reconfigure the jackson mapper or user Modules) OR you can simply create a class called MyClassDeserializer that extends JsonDeserializer and use the @JsonDeserialize(using = MyClassDeserializer.class) annotation.

example :

it's really simple and works like a charm! :)

public class User implements Identity{
     @JsonProperty("_id")
     private String id;
     @JsonDeserialize(using = OptionDeserializer.class)
     public Option<String> email;
 }

public class OptionDeserializer extends JsonDeserializer<Option> {
    @Override
    public Option deserialize(JsonParser jsonParser,
    DeserializationContext deserializationContext) throws IOException, 
    JsonProcessingException {`

        //code can be improved`

        ObjectCodec oc = jsonParser.getCodec();
        JsonNode node = oc.readTree(jsonParser);
        return  Option.apply(node.get("email").getText());
   }
}

jongo / jackson deserialize scala.option in java

1 Answers1