Question

鉴于java.util.UUID生成了这样的......

import java.util.UUID

val uuid = UUID.randomUUID

...是否可以将其转换为MongoDB ObjectID并保留唯一性？或者我应该将_id设置为UUID值？

当然，最好的解决方案是使用BSONObjectID.generate ...但在我的情况下，我得到一个 JSON Web令牌，其中UUID作为其ID，我需要跟踪它在MongoDB集合中。

的Tx。

Answer 1

为什么不直接保存_id = uuid？

MongoDB将简单地处理它：）

Answer 2

以下隐式对象允许reactivemongo将UUID转换为BSONBinary。

  implicit val uuidBSONWriter: BSONWriter[UUID, BSONBinary] =
    new BSONWriter[UUID, BSONBinary] {
      override def write(uuid: UUID): BSONBinary = {
        val ba: ByteArrayOutputStream = new ByteArrayOutputStream(16)
        val da: DataOutputStream = new DataOutputStream(ba)
        da.writeLong(uuid.getMostSignificantBits)
        da.writeLong(uuid.getLeastSignificantBits)
        BSONBinary(ba.toByteArray, Subtype.OldUuidSubtype)
      }
    }

  implicit val uuidBSONReader: BSONReader[BSONBinary, UUID] =
    new BSONReader[BSONBinary, UUID] {
      override def read(bson: BSONBinary): UUID = {
        val ba = bson.byteArray
        new UUID(getLong(ba, 0), getLong(ba, 8))
      }
    }

  def getLong(array:Array[Byte], offset:Int):Long = {
    (array(offset).toLong & 0xff) << 56 |
    (array(offset+1).toLong & 0xff) << 48 |
    (array(offset+2).toLong & 0xff) << 40 |
    (array(offset+3).toLong & 0xff) << 32 |
    (array(offset+4).toLong & 0xff) << 24 |
    (array(offset+5).toLong & 0xff) << 16 |
    (array(offset+6).toLong & 0xff) << 8 |
    (array(offset+7).toLong & 0xff)
  }

以下是使用上述编写器和读取器对象的示例

abstract class EntityDao[E <: Entity](db: => DB, collectionName: String) {

  val ATTR_ENTITY_UUID = "entityUuid"

  val collection = db[BSONCollection](collectionName)

  def ensureIndices(): Future[Boolean] = {
    collection.indexesManager.ensure(
      Index(Seq(ATTR_ENTITY_UUID -> IndexType.Ascending),unique = true)
    )
  }

  def createEntity(entity: E) = {
    val entityBSON = BSONDocument(ATTR_ENTITY_UUID -> entity.entityUuid)
    collection.insert(entityBSON)
  }

  def findByUuid(uuid: UUID)(implicit reader: BSONDocumentReader[E]): Future[Option[E]] = {
    val selector = BSONDocument(ATTR_ENTITY_UUID -> uuid)
    collection.find(selector).one[E]
  }
}

Answer 3

这是UUID的替代BSON处理程序实现，没有手动位错误：

import reactivemongo.bson._
import java.nio.ByteBuffer
import java.util.UUID

implicit val uuidBSONHandler: BSONHandler[BSONBinary, UUID] = BSONHandler(
  { bson =>
    val lb = ByteBuffer.wrap(bson.byteArray).asLongBuffer
    new UUID(lb.get(0), lb.get(1))
  },
  { uuid =>
    val arr = Array.ofDim[Byte](16)
    ByteBuffer.wrap(arr).asLongBuffer.put(uuid.getMostSignificantBits).put(uuid.getLeastSignificantBits)
    BSONBinary(arr, Subtype.UuidSubtype)
  })

如何将UUID转换为MongoDB oid

3 个答案: