CODEWITHSUNDEEP
scalaplayframeworkplayframework-2.0
Felix
Felix

Reputation: 5629

playframwork controller return ID of created object

in my scala playframework application I want to return the ID of the project I created.

controller code

  def createClient = Action { implicit request =>
    request.body.asJson.map(_.validate[ClientModel] match {
      case JsSuccess(client, _) =>
        clientDTO.createClient(client).map{
          case cnt => println(cnt)
          case  _ => println("Fehler")
        }
      case err@JsError(_) => BadRequest("TEST")
      case _ => BadRequest("fail to create Counter")
    }).getOrElse(BadRequest("Failure tu create Counter"))
    Ok("s")
  }

DTO Code

  /**
    * Insert Query for a new Client
    */
  val insertClientQuery = clients returning clients.map(_.id) into ((client, id) => client.copy(id = Some(id)))

  /**
    * Creates a new client
    *
    * @param client Client Model
    * @return
    */
  def createClient(client: ClientModel): Future[ClientModel] = {
    val action = insertClientQuery += client
    db.run(action)
  }

what would be a best preactise to the ID instead of return "OK"

thanks

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New Error enter image description here

New Ones enter image description here

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Upvotes: 0

Views: 66

Answers (1)

sheunis
sheunis

Reputation: 1544

Something like this (this will return a JSON object). You can also potentially return the whole client as a JSON object. Also, HTTP 201 (created) is a more correct response to use in this case.

def createClient = Action.async { implicit request =>
  request.body.asJson.map(_.validate[ClientModel]) match {
    case c: JsSuccess[ClientModel] =>
      clientDTO.createClient(c.get).map{
        cnt => Created(Json.obj("id" -> cnt.id))
      }.recover {
        case e: Exception => BadRequest("Could not create client")
      }
    case err: JsError => Future.successful(BadRequest("TEST"))
  }
}

Upvotes: 1

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