Jackson JSON Deserialization of MongoDB ObjectId

Question

Ok, so first off here's the JSON that's returning from my web service. I'm trying to deserialize it into pojos after an asynchronous query in a ResponseHandler in my Android ContentProvider.

{"exampleList" : [{
"locationId" : "00001" , 
"owners" : [ 
  { 
    "paidID" : { "$oid" : "50a9c951300493f64fbffdb6"} , 
    "userID" : { "$oid" : "50a9c951300493f64fbffdb6"}
  } , 
  { 
    "paidID" : { "$oid" : "50a9c951300493f64fbffdb7"} , 
    "userID" : { "$oid" : "50a9c951300493f64fbffdb7"}
  } 
]
}]}

At first, I was confused about the problem I was seeing, since I use the same Jackson-annotated beans for my web service as I do in my Android app--but then I realized that the owners object was never getting sent in the sample JSON to my web service (it skips the POJOs on my web service and gets added into the documents in mongoDB through atomic updates from the DAO).

So OK. Up to now, Jackson wasn't having to handle the owners object, and now that it is it is choking on it, namely:

JsonMappingException : Can not deserialize instance of java.lang.String out of START_OBJECT token at [char position where you can find "userID" and "paidID" ] through reference chain [path to my Jackson bean which contains the owners class]

My Jackson bean has a wrapper, which is what that "exampleList" is all about:

public class Examples extends HashMap<String, ArrayList<Example>> {

}

And then the actual Example class:

@JsonIgnoreProperties(ignoreUnknown = true)
public class Example implements Comparable<Example> {

@ObjectId @Id
private String id;

@JsonProperty(Constants.Example.location)
private String location;

@JsonProperty(Constants.Example.OWNERS)
private List<Owners> owners;

public int compareTo(Example _o) {
    return getId().compareTo(_o.getId());
}

public String getId() {
    return id;
}

public void setId(String id) {
    this.id = id;
}

public String getLocation() {
    return location;
}

public void setLocation(String location) {
    this.location = location;
}
public List<Example.Owners> getOwners() {
    return owners;
}

public void setOwners(List<Example.Owners> owners) {
    this.owners = owners;
}

public Example() {
}

@JsonCreator
public Example(@Id @ObjectId String id) {
    this.id = id;
}

@JsonIgnoreProperties(ignoreUnknown = true)
public static class Owners implements Comparable<Owners> {

    @JsonProperty(Constants.Example.USERID)
    private String userID;

    @JsonProperty(Constants.Example.PAIDID)
    private String paidID;

    public Owners() {
    }

    public int compareTo(Owners _o) {
        return getUserID().compareTo(_o.getUserID());
    }

    @ObjectId
    public String getUserID() {
        return userID;
    }

    @ObjectId
    public void setUserID(String userID) {
        this.userID = userID;
    }

    @ObjectId
    public String getPaidID() {
        return paidID;
    }

    @ObjectId
    public void setPaidID(String paidID) {
        this.paidID = paidID;
    }

}

}

And finally, the code in the ResponseHandler where this is all failing (the 2nd line produces the JsonMappingException):

objectMapper = MongoJacksonMapperModule.configure(objectMapper);    
mExamples = objectMapper.readValue(jsonParser, Examples.class);

I have a feeling the issue is that Jackson still doesn't know how to map those $oid , which are the mongoDB ObjectIds. The MongoJacksonMapper library is supposed to help that by providing the @ObjectId annotation and a way to configure the ObjectMapper to use that library, but it still isn't working. For some reason, it's still looking for the userID or paidID as a String, not an ObjectId. Any ideas?

Answer 1

Another alternative is com.fasterxml.jackson.databind.ser.std.ToStringSerializer .

@Id
@JsonSerialize(using = ToStringSerializer.class)
private final ObjectId id;

This will result in:

{
  "id": "5489f420c8306b6ac8d33897"
}

Answer 2

For future users: Use a custom jackson deserializer to convert $oid back to ObjectId.

public class ObjectIdDeserializer extends JsonDeserializer<ObjectId> {

    @Override
    public ObjectId deserialize(JsonParser p, DeserializationContext ctxt) throws IOException, JsonProcessingException {
        JsonNode oid = ((JsonNode)p.readValueAsTree()).get("$oid");
        return new ObjectId(oid.asText());
    }

}

How to use:

ObjectMapper mapper = new ObjectMapper();
SimpleModule mod = new SimpleModule("ObjectId", new Version(1, 0, 0, null, null, null));
        
mod.addDeserializer(ObjectId.class, new ObjectIdDeserializer());
mapper.registerModule(mod);

YourClass obj = mapper.readValue("{your json with $oid}", YourClass.class);

Answer 3

My code had at least two problems that were pretty tough to track down answers to online, so I'll make sure to link here. Basically, child classes need a constructor in the parent class that calls Jackson's readValue() to map the child. As far as mongoDB $oid's go, you should create a separate MongoId class to represent these mongo objects, and follow a similar pattern as with the child class to map the data when it comes in for deserialization. Here's a blog post I found that describes this well and provides some examples.

Answer 4

Jackson does not know how to serialize an ObjectId. I tweaked Arny's code to serialize any ObjectId and provide this working example:

public class SerialiserTest {

 private ObjectMapper mapper = new ObjectMapper();

 public static class T {
    private ObjectId objectId;

    public ObjectId getObjectId() {
        return objectId;
    }

    public void setObjectId(ObjectId objectId) {
        this.objectId = objectId;
    }
 }

 @Test
 public final void serDeser() throws IOException {
    T t = new T();
    t.setObjectId(new ObjectId());
    List<T> ls = Collections.singletonList(t);
    String json = mapper.writeValueAsString(ls);
    System.out.println(json);
    SimpleModule mod = new SimpleModule("ObjectId", new Version(1, 0, 0, null, null, null));
    mod.addDeserializer(ObjectId.class, new ObjectIdDeserializer());
    mapper.registerModule(mod);
    JavaType type = mapper.getTypeFactory().
            constructCollectionType(List.class, T.class);
    List<?> l  = mapper.readValue(json, type);
    System.out.println(l);
 }
}

public class ObjectIdDeserializer extends JsonDeserializer<ObjectId> {

 @Override
 public ObjectId deserialize(JsonParser p, DeserializationContext ctxt) throws IOException, JsonProcessingException {
    JsonNode n = (JsonNode)p.readValueAsTree();
    return new ObjectId(n.get("timestamp").asInt(), n.get("machineIdentifier").asInt(), (short) n.get("processIdentifier").asInt(), n.get("counter").asInt());
 }

}

Answer 5

There's an even easier way documented here which was a lifesaver for me. Now you can use the ObjectId in Java but when you go to/from JSON it'll be a String.

public class ObjectIdJsonSerializer extends JsonSerializer<ObjectId> {
    @Override
    public void serialize(ObjectId o, JsonGenerator j, SerializerProvider s) throws IOException, JsonProcessingException {
        if(o == null) {
            j.writeNull();
        } else {
            j.writeString(o.toString());
        }
    }
}

And then in your beans:

@JsonSerialize(using=ObjectIdJsonSerializer.class)
private ObjectId id;

Answer 6

I did it like this:

@Configuration
public class SpringWebFluxConfig {

    @Bean
    @Primary
    ObjectMapper objectMapper() {
      Jackson2ObjectMapperBuilder builder = new Jackson2ObjectMapperBuilder();
      builder.serializerByType(ObjectId.class, new ToStringSerializer());
      builder.deserializerByType(ObjectId.class, new JsonDeserializer() {
        @Override
        public Object deserialize(JsonParser p, DeserializationContext ctxt) 
            throws IOException {
          Map oid = p.readValueAs(Map.class);
          return new ObjectId(
              (Integer) oid.get("timestamp"),
              (Integer) oid.get("machineIdentifier"),
              ((Integer) oid.get("processIdentifier")).shortValue(),
              (Integer) oid.get("counter"));
        }
      });
      return builder.build();
    }
}