当其中一个构造函数参数默认初始化时,如何从Json创建实例?

当其中一个构造函数参数默认初始化时,如何从Json创建实例?,json,mongodb,scala,spray,spray-json,Json,Mongodb,Scala,Spray,Spray Json,ADSRegistrationMap中的方法用于从MongoDB保存和检索文档。ObjectId是在初始化期间创建的。我也必须这样做才能从Json加载注册,Json是POST正文的一部分,所以我想我可以添加ADSRegistrationProtocol对象来实现这一点。它因编译错误而失败。有没有办法解决这个问题或者做得更好 package model import spray.json._ import DefaultJsonProtocol._ import com.mongodb.casb

ADSRegistrationMap中的方法用于从MongoDB保存和检索文档。ObjectId是在初始化期间创建的。我也必须这样做才能从Json加载注册,Json是POST正文的一部分,所以我想我可以添加ADSRegistrationProtocol对象来实现这一点。它因编译错误而失败。有没有办法解决这个问题或者做得更好

package model

import spray.json._
import DefaultJsonProtocol._
import com.mongodb.casbah.Imports._
import org.bson.types.ObjectId
import com.mongodb.DBObject
import com.mongodb.casbah.commons.{MongoDBList, MongoDBObject}

case class Registration(
  system: String, 
  identity: String, 
  id: ObjectId = new ObjectId())

object RegistrationProtocol extends DefaultJsonProtocol {
  implicit val registrationFormat = jsonFormat2(Registration)
}

object RegistrationMap {
  def toBson(registration: Registration): DBObject = {
    MongoDBObject(
      "system"         -> registration.system,
      "identity"       -> registration.identity,
      "_id"            -> registration.id
    )
  }

  def fromBson(o: DBObject): Registration = {
    Registration(
      system = o.as[String]("system"),
      identity = o.as[String]("identity"),
      id = o.as[ObjectId]("_id")
    )
  }
}
编译错误:

[error] /model/Registration.scala:20: type mismatch;
[error]  found   : model.Registration.type
[error]  required: (?, ?) => ?
[error]  Note: implicit value registrationFormat is not applicable here because it comes after the application point and it lacks an explicit result type
[error]   implicit val registrationFormat = jsonFormat2(Registration)
[error]                                                    ^
[error] one error found
[error] (compile:compile) Compilation failed


将ObjectId更新为String,将jsonFormat2更新为jsonFormat3以修复编译错误

case class Registration(
  system: String, 
  identity: String, 
  id: String = (new ObjectId()).toString())

object RegistrationProtocol extends DefaultJsonProtocol {
  implicit val registrationFormat = jsonFormat3(Registration)
}
将POST请求主体转换为注册对象时立即获取运行时错误。有什么想法吗

val route: Route = {
  pathPrefix("registrations") {
    pathEnd {
      post {
        entity(as[Registration]) { registration =>
下面是build.sbt中的内容

scalaVersion := "2.10.4"

scalacOptions ++= Seq("-feature")

val akkaVersion = "2.3.8"

val sprayVersion = "1.3.1"

resolvers += "spray" at "http://repo.spray.io/"
resolvers += "Sonatype releases" at "https://oss.sonatype.org/content/repositories/releases"

// Main dependencies
libraryDependencies ++= Seq(
    "com.typesafe.akka" %% "akka-actor" % akkaVersion,
    "com.typesafe.akka" %% "akka-slf4j" % akkaVersion,
    "com.typesafe.akka" %% "akka-camel" % akkaVersion,
    "io.spray" % "spray-can" % sprayVersion,
    "io.spray" % "spray-routing" % sprayVersion,
    "io.spray" % "spray-client" % sprayVersion,
    "io.spray" %% "spray-json" % sprayVersion,
    "com.typesafe" % "config" % "1.2.1",
    "org.apache.activemq" % "activemq-camel" % "5.8.0",
    "ch.qos.logback" % "logback-classic" % "1.1.2",
    "org.mongodb" %% "casbah" % "2.7.4"
)
错误:

12:33:03.477 [admcore-microservice-system-akka.actor.default-dispatcher-3] DEBUG s.can.server.HttpServerConnection - Dispatching POST request to http://localhost:8878/api/v1/adsregistrations to handler Actor[akka://admcore-microservice-system/system/IO-TCP/selectors/$a/1#-1156351415]
Uncaught error from thread [admcore-microservice-system-akka.actor.default-dispatcher-3] shutting down JVM since 'akka.jvm-exit-on-fatal-error' is enabled for ActorSystem[admcore-microservice-system]
java.lang.NoSuchMethodError: spray.json.JsonParser$.apply(Ljava/lang/String;)Lspray/json/JsValue;
    at spray.httpx.SprayJsonSupport$$anonfun$sprayJsonUnmarshaller$1.applyOrElse(SprayJsonSupport.scala:36)
    at spray.httpx.SprayJsonSupport$$anonfun$sprayJsonUnmarshaller$1.applyOrElse(SprayJsonSupport.scala:34)

为了避免任何问题,我将如下定义Register类(它似乎是一个数据模型

案例类寄存器(系统:字符串,标识:字符串,标识:字符串)
这是因为将
id
字段作为字符串而不是BSON ObjectId(我习惯于使用不依赖第三方库的数据模型)对我来说更有意义

因此,正确的JSON协议将使用
jsonFormat3
,而不是
jsonFormat2

对象注册协议扩展了DefaultJsonProtocol{
隐式val registrationFormat=jsonFormat3(注册)
}
这将解决任何类型的JSON序列化问题

最后,您的
toBson
fromBson
转换器将是:

def toBson(r:注册):DBObject={ MongoDBObject( “系统”->r系统, “身份”->r.identity, “\u id”->新对象id(r.id) ) } 及

def fromBson(o:DBObject):注册={
登记(
系统=o.as[字符串](“系统”),
identity=o.as[String](“identity”),
id=o.as[ObjectId](“_id”).toString
)
}

A这就是使用BSON ObjectId的地方:更接近于依赖MongoDB的逻辑。

我认为这与相关,因为默认情况下,
ObjectId
不可序列化。此外,您可能需要将
jsonFormat2
更改为
jsonFormat3
将ObjectId更新为String,将jsonFormat2更改为jsonFormat3以修复编译错误。现在,将POST请求主体转换为注册实例时出现新的运行时错误。有什么想法吗?是的,我认为您遇到了spray json的兼容性问题,请参阅。降级到
1.2.6
,事情就会解决。或者,您也可以复制通知单中描述的
SprayJsonSupport
。降级到1.2.6有效。谢谢