Future and Scala: Data type mismatch

I have the next code:

package controllers.usersPage

import play.api.mvc._
import play.api.libs.json._
import model.{User, Users, Patients, CObject}
import scala.concurrent.Future

import service.{UserService}

class allUsersToJSON() extends Controller {

def convertUsersToJsonOrig(lusers: Seq[User]): JsValue = {Json.toJson(
                                           lusers.map { u => Map("id" -> u.id, "firstName" -> u.firstName, "lastName" -> u.lastName, "email" -> u.email, "username" -> u.username, "password" -> u.password)}) }


def retAllUsers = Action { request =>
   Ok( Json.stringify(convertUsersToJsonOrig(Users.listAll)))
}

}

But I am having the next error:

$ compile
[info] Compiling 69 Scala sources and 3 Java sources to
H:\project\target\scala-2.11\classes…
[error] H:\project\app\controllers\usersPage\retrieveAllUsersJSON.scala:29: type mismatch;
[error] found : scala.concurrent.Future[Seq[model.User]]
[error] required: Seq[model.User]
[error] Ok( Json.stringify(convertUsersToJsonOrig(Users.listAll)))
[error] ^
[warn] Class com.sun.tools.xjc.Options not found - continuing with a stub.
[warn] one warning found
[error] one error found
[error] (root/compile:compileIncremental) Compilation failed
[error] Total time: 7 s, completed Apr 17, 2017 10:51:11 AM

The instruction Users.listAll is an Future one that will give me a list of User objects that I will need to convert to JSon. As I see this command gives me the scala.concurrent.Future[Seq[model.User]] data type. How can I do to get the Seq[model.user] from this command so I can use it within convertUsersToJsonOrig? Thanks

Thanks

Hello,

You probably want to use Play asynchronously
https://www.playframework.com/documentation/2.5.x/ScalaAsync.

 Best, Oliver