Question

我有以下代码

package controllers

import play.api.mvc._

/**
  * Provide security features
  */
trait Secured {

  /**
    * Retrieve the connected user's email
    */
  private def username(request: RequestHeader) = request.session.get("username")

  /**
    * Not authorized, forward to login
    */
  private def onUnauthorized(request: RequestHeader) = {
    Results.Redirect(routes.Authentication.index)
  }

  /**
    * Action for authenticated users.
    */
  def IsAuthenticated(f: => String => Request[AnyContent] => Result) = {
    Security.Authenticated(username, onUnauthorized) { user =>
      Action(request => f(user)(request))
    }
  }
}

还有

package controllers

import javax.inject.Inject
import play.api.i18n.MessagesApi
import play.api.mvc.Controller
import service.userService
import views._
import scala.concurrent.ExecutionContext.Implicits.global

class Restricted @Inject()
(userService: userService,
 val messagesApi: MessagesApi) extends Controller with Secured {

  /**
    * Display restricted area only if user is logged in.
    */
  def index = IsAuthenticated { user =>
    _ => userService.getUser(username).map {
      Ok(views.html.restricted(user))
    }.getOrElse(Forbidden)
  }

}

但是我无法编译该代码，它在类型不匹配的views.html.restricted（user）上持续失败，应该是：字符串=> Request [AnyContent] =>结果，实际是：String => Request [AnyContent] = >任何。我是scala和游戏的超级新手，经过数小时的谷歌搜索之后，我找不到与我的错误类似的东西。

Answer 1

我设法通过将userService.getUser的返回类型从选项[User]更改为User并添加try / catch来处理（如果找不到）来解决此问题。

无法解决简单身份验证Scala中的类型错误，播放框架项目

1 个答案: