使用 scala mongo 驱动程序序列化为对象? [英] Serialize to object using scala mongo driver?

问题描述

我是新来斯卡拉蒙戈司机和想了解如何映射从文档类?似乎没有任何文档显示这是如何完成的.在 .net 驱动程序中，它就像传递泛型和自动映射字段一样简单.scala 中没有类似的吗?

I'm new to the scala mongo driver and am trying to understand how to map a class from a Document? None of the documentation seems to show how this is done. In the .net driver, its as easy as passing a generic and having fields auto mapped. Is there nothing similar in scala?

推荐答案

它们并不容易.通过java挖掘，我想出了这个解决方案:

They don't make it easy. Digging through the java, I came up with this solution:

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")
}

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