Scala Play 2.4如何测试动作组合(使用动作构建器和动作过滤器)
发布时间:2020-12-16 19:23:16 所属栏目:安全 来源:网络整理
导读:我编写了一个动作构建器,并使用Injection将此新动作应用于我的控制器. 以下是我的代码: class AuthAction @Inject()(authService: AuthService) { def AuthAction(userId: Int) = new ActionBuilder[Request] with ActionFilter[Request] { def filter[A](r
我编写了一个动作构建器,并使用Injection将此新动作应用于我的控制器.
以下是我的代码: class AuthAction @Inject()(authService: AuthService) { def AuthAction(userId: Int) = new ActionBuilder[Request] with ActionFilter[Request] { def filter[A](request: Request[A]): Future[Option[Result]] = { request.headers.get("Authorization").map(header => { authService.isAuth(header).flatMap { case true => Future.successfuly(None) case false => Future.successfuly(Some(Forbidden)) } }) } } } class UserController @Inject()(auth: AuthAction,xxxService: XxxService) extends Controller { def xxx(userId: Int) = auth.AuthAction(userId).async { implicit request => xxxService.xxx().map....... } } 不使用自定义操作,我可以使用ScalaTest和Mockito轻松测试UserController(假请求和模拟xxxService) 我的问题是如何使用ScalaTest和Mockito模拟AuthAction并测试UserController? 谢谢 解决方法val authAction = mock[AuthAction] when(authAction.AuthAction(any[Int])).thenReturn { new ActionBuilder[Request] with ActionFilter[Request] { def filter[A](request: Request[A]): Future[Option[Result]] = Future.successful(None) /* or Future.successful(Some(Forbidden)) */ } } (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |