播放Scala:类需要是抽象的
我是Scala游戏的初学者,当我想打开我的应用程序时,我会收到以下错误:播放Scala:类需要是抽象的,scala,playframework,Scala,Playframework,我是Scala游戏的初学者,当我想打开我的应用程序时,我会收到以下错误: class ElevesController needs to be abstract, since method messagesApi in trait I18nSupport of type => play.api.i18n.MessagesApi is not defined 电子控制器 package controllers import javax.inject.Inject import play.
class ElevesController needs to be abstract, since method messagesApi in trait I18nSupport of type => play.api.i18n.MessagesApi is not defined
电子控制器
package controllers
import javax.inject.Inject
import play.api.Logger
import play.api.data.Form
import play.api.data.Forms._
import play.api.i18n.{Messages, I18nSupport, MessagesApi}
import play.api.mvc.{Action, Controller}
import views.html
import scala.concurrent.Future
import models.Eleve
class ElevesController @Inject() extends Controller with I18nSupport {
def viewEleves = Action {implicit request =>
Ok(html.viewseleves(Eleve.findAll))
}
}
我不知道如何修理它
编辑:
我修复了它,我只需要在这行中添加:“(val messagesApi:messagesApi)”:
class ElevesController @Inject()(val messagesApi: MessagesApi) extends Controller with I18nSupport {
这应该行得通。只要改变你的行为等等
嗯,出于某种原因,我们一直在使用
@Inject()(messagesApi:messagesApi)
而没有编译的val
,直到你需要混音!
package controllers
import javax.inject._
import play.api._
import javax.inject.Inject
import play.api.Logger
import play.api.data.Form
import play.api.data.Forms._
import play.api.mvc._
import views.html
import scala.concurrent.Future
import play.api.i18n._
@Inject()
class ElevesController (val messagesApi: MessagesApi) extends Controller with I18nSupport {
val action = Action { implicit request =>
Ok("Ok") // Uses the client???s preferred language
}
}