Mongodb 使用scala mongo驱动程序序列化到对象？

我是scala mongo驱动程序的新手，正在尝试了解如何从文档映射类？所有文档似乎都没有说明这是如何实现的。在.net驱动程序中，只需传递泛型并自动映射字段即可。scala中没有类似的东西吗？

它们不容易做到。通过深入研究java，我提出了以下解决方案：

import org.bson.codecs.DecoderContext
import org.bson.codecs.configuration.CodecRegistries.{fromProviders, fromRegistries}
import org.bson.codecs.configuration.CodecRegistry
import org.bson.{BsonDocumentReader, BsonDocumentWrapper}
import org.mongodb.scala.bson.codecs.{DEFAULT_CODEC_REGISTRY, Macros}
import org.mongodb.scala.bson.collection.mutable.Document

import scala.reflect.classTag

case class Person(firstName: String, lastName: String)

object MongoTest extends App {

  val personCodecProvider = Macros.createCodecProvider[Person]()
  val codecRegistry: CodecRegistry = fromRegistries(fromProviders(personCodecProvider), DEFAULT_CODEC_REGISTRY)

  val document = Document("firstName" -> "first", "lastName" -> "last")
  val bsonDocument = BsonDocumentWrapper.asBsonDocument(document, DEFAULT_CODEC_REGISTRY)

  val bsonReader = new BsonDocumentReader(bsonDocument)
  val decoderContext = DecoderContext.builder.build
  val codec = codecRegistry.get(classTag[Person].runtimeClass)
  val person: Person = codec.decode(bsonReader, decoderContext).asInstanceOf[Person]

  println(s"person: $person")
}

---

使用mongo宏处理程序序列化和反序列化对象的示例

import reactivemongo.api.bson.{BSON, BSONDocument, Macros}

case class Person(name:String = "SomeName", age:Int = 20)

implicit val personHandler = Macros.handler[Person]

//Serialize
val bsonPerson = BSON.writeDocument[Person](Person())

println(s"${BSONDocument.pretty(bsonPerson.getOrElse(BSONDocument.empty))}")

//Deserialize

val bsonDocumentPerson = BSONDocument("name"-> "MyNameHere", "age"->35)

val scalaObjPerson: Person = BSON.read[Person](bsonDocumentPerson).getOrElse(Person())

printf(s"Scala person obj = $scalaObjPerson")

---

org.mongodb.scala.bson.codecs.DEFAULT\u CODEC\u注册表

已弃用，请使用

import MongoClient.DEFAULT\u CODEC\u注册表

。棒极了，格雷德！