Return original response from webservice call on scala with play

Question

I have have a scala project (V. 2.12.12) witch use play framework and i try to call webservice and return response.

To do that, i create a client :

  override def postStatus(body: JsValue): Future[WSResponse] = {
    wsClientHelper.callWS(updateStatusUrl, POST, callerBody = body)
  }

My method return Future[WSResponse]. So i call them in a controller :

  def postKycStatus: Action[AnyContent] = action.async { implicit request =>
    kycRepository
      .postStatus(request.body.asJson.getOrElse(JsObject.empty))
  }

But when i do that, i can call the updateStatus api sucessfully with postStatus but i got an compilation error :

play.sbt.PlayExceptions$CompilationException: Compilation error[type mismatch; found : scala.concurrent.Future[play.api.libs.ws.WSResponse] required: scala.concurrent.Future[play.api.mvc.Result]]

How can i return my original webservice response with original status code ?

Thanks to your help !

Answer 1

I found the response. You can do it with:

def postKycStatus: Action[AnyContent] = action.async { implicit request =>
  kycRepository
  .postStatus(request.body.asJson.getOrElse(JsObject.empty))
  .map(response => Status(response.status)(response.body).as(JSON))
}