Question

我有这个方法：

def withAuth(f: => User => Request[AnyContent] => Result) = {
    Authentication.isAuthenticated(AuthenticationToken(AuthenticationService.TokenKey)) match {
      case None => Results.Redirect(routes.AuthenticationService.notLoggedIn)
      case Some(user) => Action(request => f(user)(request))
    }
  }

我用它像：

  def list(locationId: Option[Int]) = withAuth { user =>
    implicit request =>
      val entities = Assets.filter(user, locationId)
      Logger.info("Succesfully returned %d assets to user %s".format(entities.length, user))
      Ok(Json.toJson(entities.map(s => Json.toJson(s))))
  }

正如您所注意到的，我希望将其用作一种方法，如果用户未登录，则Redirect将他登录页面，否则会从会话返回用户。问题在于Redirect，在运行时Play正在抱怨：

不能使用将Object作为处理程序返回的方法

有人有任何线索吗？

Answer 1

为了避免上述问题，我最终做到了：

   def onUnauthorized(request: RequestHeader) = Results.Redirect(routes.AuthenticationService.notLoggedIn)

  /**
   * A very important wrapper method which checks if the user is logged-in: if it is, return the User, otherwise
   * redirect the user to a specific page.
   */
  def withAuthentication(f: => Option[User] => Request[AnyContent] => Result) = {
    Security.Authenticated(userId, onUnauthorized) { user =>
      Action(request => f(Users.findById(Integer.valueOf(user)))(request))
    }
  }

Answer 2

我认为withAuth的返回类型不是你想象的那样。我认为您需要将Redirect包装在Action中。

为什么我在Play 2中收到特定方法的错误？

2 个答案: