playframework scala returning post request - scala

Question

in my scala playframework application I want to return the via Post submitted and then with slick stored object back to frontend as json
I tried this:
def createClient = Action.async { implicit request =>
request.body.asJson.map(_.validate[ClientModel] match {
case JsSuccess(client) =>
clientDTO.createClient(client).map { clients =>
Ok(Json.toJson(clients))
}
})
}
but I get this error:
what could be my problem?
NEW ERROR

Answer 1

Try with something along these lines:
def createClient = Action.async { implicit request =>
request.body.asJson match {
case None => // do something that returns a Future[Result] ~ such as NotFound or
case Some(js) =>
js.validate[ClientModel] match {
case client: JsSuccess[ClientModel] =>
clientDTO.createClient(client).map { clients =>
Ok(Json.toJson(clients))
}
case e: JsError => // do something that returns a Future[Result] ~ such as InternalServerError
}
}
}

Answer 2

the .async, as the name suggests require a Future type.
You have 2 options:
Remove the .async, this will make your def synchronous (deprecated)
Leave the .async but return a Future result
def createClient = Action.async { implicit request =>
request.body.asJson.map(_.validate[ClientModel] match {
case JsSuccess(client) =>
clientDTO.createClient(client).map { clients =>
Future(Ok(Json.toJson(clients)))
}
})
}
But you still need to add the case of validation error:
case JsSuccess(client) =>
{
clientDTO.createClient(client).map
{
clients => Future(Ok(Json.toJson(clients)))
}
}
case _ => Future(BadRequest(""))
This should work and, in all the cases apart JsSuccess, the function will return a future BadRequest response.
A better solution is to change the _ with JsError:
case e: JsError =>
{
Println(e)
Future(BadRequest(.....))
}
This will also print the error.
You can read more here: https://www.playframework.com/documentation/2.6.x/ScalaJson
(Using validation chapter)
More about future in scala: https://docs.scala-lang.org/overviews/core/futures.html