Scala Play Sample App Json Serializer errors

Question

I am trying Scala with Play framework and coming from 10+ years of Java/Spring experience

Below is the source code I am trying

Routes file

GET    /stock                      controllers.HomeController.saveStock

Sample Model object

import play.api.libs.json._

case class Stock(symbol: String = "", price: Double = 0d)

object Stock  {
  implicit def stockReads = Json.reads[Stock]
  implicit def stockWrites = Json.writes[Stock]
  implicit def stockFormat = Json.format[Stock]
}

Sample Controller

import models.Stock
import javax.inject._
import play.api.mvc._
import play.api.libs.json._

@Singleton
class HomeController @Inject()(cc: ControllerComponents) extends AbstractController(cc: ControllerComponents) {
  def index() = Action { implicit request: Request[AnyContent] =>
    Ok(views.html.index())
  }
    def saveStock = Action{ implicit request: Request[AnyContent] =>
    var stocks = List(Stock())
    println(stocks)
    Ok(Json.toJson(Stock)).as("application/json")
    //Ok
  }
}

build.sbt

scalaVersion := "2.13.6"

libraryDependencies += guice
libraryDependencies += "org.scalatestplus.play" %% "scalatestplus-play" % "5.0.0" % Test
libraryDependencies += "com.typesafe.play" %% "play-server" % "2.8.8"
libraryDependencies += "com.typesafe.play" %% "play-slick" % "5.0.0"
libraryDependencies += "com.github.tminglei" %% "slick-pg" % "0.19.7"
libraryDependencies += "ai.x" %% "play-json-extensions" % "0.42.0"

Error - http://localhost:9000/ No Json serializer found for type models.Stock.type. Try to implement an implicit Writes or Format for this type.

Gaël J Gaël J · Accepted Answer · 2021-08-14T16:35:33

In the error message "No Json serializer found for type models.Stock.type", one important thing to notice is the .type suffix.

It's referring to the object Stock, not the case class. That is the error is saying "I don't know how to serialize object Stock".

This is because you wrote Json.toJson(Stock) instead of Json.toJson(Stock()) or maybe you meant Json.toJson(stocks).

Stock refers to the object while Stock() is Instantiating the case class.

Also:

you don't need to specify the as(...) if using a JsObject in the result.
you dont'e need a Reads nor a Writes if you provide a Format (the later is providing the 2 former)
use val rather than var (immutability is the strength of Scala)

Scala Play Sample App Json Serializer errors

1 Answers