async scala driver: how to encode an UUID in BSON

[Yann Simon]

Hi,

I am looking at using BSON: http://mongodb.github.io/mongo-scala-driver/1.1/bson/documents/

I have not found out how to encode a java.util.UUID.

By looking into the code of the different drivers, I found out I could use BsonBinary with the right BsonBinarySubType. But then I have to convert the UUID to an array of bytes myself.

Thanks for any pointer,

Yann

[Jeff Yemin]

Hi Yann,

It's not elegant, but here's one way to do it: https://gist.github.com/jyemin/45996b72977c4453ca7f.

Make sure to choose an appropriate UuidRepresentation based upon the needs of your particular application.

Regards,

Jeff

[Yann Simon]

Thank for the feedback Jeff!

I will try to incorporate that. For the moment, I created BasicDBObject and converted it with BsonDocumentWrapper.

It is not elegant neither ;)

Do you think it'd make sense to add a BsonUUID to encapsulate this logic?

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[Jeff Yemin]

I haven't considered it, but please feel free to open a ticket for that in the JAVA project in Jira.

Regards,

Jeff

[Damir Vandic]

I believe this seemed to work for me:

import java.nio.ByteBuffer
import java.util.UUID

import org.bson.{BsonBinary, BsonBinarySubType}
import org.mongodb.scala.bson
import org.mongodb.scala.bson.BsonTransformer

object Test {
  implicit val uuidTransformer: BsonTransformer[UUID] = new BsonTransformer[UUID] {
    override def apply(value: UUID): bson.BsonValue = {
      val bb = ByteBuffer.wrap(new Array[Byte](16))
      bb.putLong(value.getMostSignificantBits)
      bb.putLong(value.getLeastSignificantBits)
      new BsonBinary(BsonBinarySubType.UUID_STANDARD, bb.array)
    }
  }
}

[Yann Simon]

Thanks Damir, your approach works as well!

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