No Json序列化程序作为JsObject发现的类型reactivemongo.play.json.JSONSerializationPack.type

Question

我正在使用具有反应性的mongoDB的play framework 2.5.3。

import javax.inject._ 

import model._ 
import play.api.Logger 
import play.api.libs.json._ 
import play.api.mvc._ 
import play.modules.reactivemongo._ 
import reactivemongo.api.ReadPreference 
import reactivemongo.play.json._ 
import reactivemongo.play.json.collection._ 

import scala.concurrent.{ExecutionContext, Future} 


class InsertController @Inject()(val reactiveMongoApi: ReactiveMongoApi)(implicit exec: ExecutionContext) extends Controller with MongoController with ReactiveMongoComponents { 

def dataFuture: Future[JSONCollection] = database.map(_.collection[JSONCollection]("data")) 

def createFromJson = Action.async(parse.json) { request => 
Json.fromJson[jsonWrapper](request.body) match { 
    case JsSuccess(data, _) => 
    for { 
     data <- dataFuture 
     lastError <- data.insert(data) 
    } yield { 
     Logger.debug(s"Successfully inserted with LastError: $lastError") 
     Ok("Inserted into db") 

    } 
    case JsError(errors) => 
    Future.successful(BadRequest("Something went wrong")) 
} 
} 

这是我的控制器和编译时，我得到以下异常：

[info] Compiling 6 Scala sources and 1 Java source to /home/***/target/scala-2.11/classes... 
[error] /home/***/app/controllers/InsertController.scala:38: No Json serializer as JsObject found for type reactivemongo.play.json.JSONSerializationPack.type. Try to implement an implicit OWrites or OFormat for this type. 
[error]   lastError <- data.insert(data.pack) 
[error]         ^
[error] one error found 
[error] (compile:compile) Compilation failed 
[info] Compiling 6 Scala sources and 1 Java source to /home/***/target/scala-2.11/classes... 
[error] /home/***/app/controllers/InsertController.scala:38: No Json serializer as JsObject found for type reactivemongo.play.json.JSONSerializationPack.type. Try to implement an implicit OWrites or OFormat for this type. 
[error]   lastError <- data.insert(data.pack) 
[error]         ^
[error] one error found 
[error] (compile:compile) Compilation failed 
[error] application - 

! @705di1397 - Internal server error, for (GET) [/] -> 

play.sbt.PlayExceptions$CompilationException: Compilation error[No Json serializer as JsObject found for type reactivemongo.play.json.JSONSerializationPack.type. Try to implement an implicit OWrites or OFormat for this type.] 
    at play.sbt.PlayExceptions$CompilationException$.apply(PlayExceptions.scala:27) 
    at play.sbt.PlayExceptions$CompilationException$.apply(PlayExceptions.scala:27) 
    at scala.Option.map(Option.scala:145) 
    at play.sbt.run.PlayReload$$anonfun$taskFailureHandler$1.apply(PlayReload.scala:49) 
    at play.sbt.run.PlayReload$$anonfun$taskFailureHandler$1.apply(PlayReload.scala:44) 
    at scala.Option.map(Option.scala:145) 
    at play.sbt.run.PlayReload$.taskFailureHandler(PlayReload.scala:44) 
    at play.sbt.run.PlayReload$.compileFailure(PlayReload.scala:40) 
    at play.sbt.run.PlayReload$$anonfun$compile$1.apply(PlayReload.scala:17) 
    at play.sbt.run.PlayReload$$anonfun$compile$1.apply(PlayReload.scala:17) 

它建议编写Owrites或OFormat为JSONCollection这是reactivemongo.play.json._包的一部分，应该已经将这些定义为我的理解。

这里是我jsonWrapper类：

case class jsonWrapper(tables : tables, userId : String) 


object jsonWrapper{ 
    implicit val jsonRead: Reads[jsonWrapper] = (
    (JsPath \ "tables").read[tables] and 
     (JsPath \ "userID").read[String] 
    )(jsonWrapper.apply _) 

    implicit val jsonWrites: Writes[jsonWrapper] = (
    (JsPath \ "tables").write[tables] and 
     (JsPath \ "userID").write[String] 
    )(json => (json.tables, json.userId)) 

    implicit val jsonWrapperFormat : Format[jsonWrapper] = Json.format[jsonWrapper] 

} 

的表类也有implictly定义的格式，读取和写入。

我orignally用这个例子开始： https://github.com/jonasanso/play-reactive-mongo-db#master，其工作但是当我尝试以使其适应我的需要（即我jsonWrapper类）我得到这个错误，不明白为什么它不工作。

非常感谢，

彼得·

Answer 1

我发现我的错误。

def createFromJson = Action.async(parse.json) { request => 
Json.fromJson[jsonWrapper](request.body) match { 
    case JsSuccess(data, _) => 
    for { 
     data <- dataFuture 
     lastError <- data.insert(data) 
    } yield { 
     Logger.debug(s"Successfully inserted with LastError: $lastError") 
     Ok("Inserted into db") 

    } 
    case JsError(errors) => 
    Future.successful(BadRequest("Something went wrong")) 
} 

我的案例条目启动一个名为“data”的对象，我通过覆盖我的dataFuture对象来跟进。因此，导致错误。我只是不得不改变变量名...

我倒了一种愚蠢。