安全的社交 websockets 播放框架答案

【问题标题】：secure social websockets play framework安全的社交 websockets 播放框架
【发布时间】：2014-04-19 06:45:55
【问题描述】：

如何使用 scala securesocial 在 play 框架中保护 websocket 调用？

def statusfeed() = WebSocket.using[String] { implicit request =>
    if (logged in) {
      ...
    else {
    }
}

谢谢，

编辑 1：

试过了但没用，总是“未登录”

def statusfeed() = WebSocket.using[String] { implicit request =>
  var in = Iteratee.ignore[String]
  var out = Enumerator.empty[String]
  session.get("userId").map { sessionId =>
      //sessionId is now userId in session
      //check he is authorise to view page or not   
      //  Ok("user in session")
      def getLoadAverage = {
          "%1.2f" format
            ManagementFactory.getOperatingSystemMXBean.getSystemLoadAverage()
      }

      in = Iteratee.ignore[String]
      out = Enumerator.repeatM {
          Promise.timeout(getLoadAverage, 3 seconds)
      }

  }.getOrElse {
      log.info("not logged in")
      //any thing that you want when he is not in session
      Redirect("/") //NOT IN SESSION REDIRECT TO LOGIN PAGE
  }
  (in, out)
}

编辑 2：用 SecureSocial.USER_KEY 替换 userId 也不起作用

【问题讨论】：

标签： scala playframework securestring

【解决方案1】：

这可行，但不是惯用的scala

def getUser(req: RequestHeader):String = {
  val authenticator = SecureSocial.authenticatorFromRequest(req)
  if (authenticator.isEmpty) {
    return null
  } 
  val auth = UserService.find(authenticator.get.identityId)
  auth.get.identityId.userId
}

def statusfeed() = WebSocket.using[String] { implicit request =>
    implicit val req = request
    val user = getUser(req)

....

然后测试用户！= null

或者更简单

def statusfeed() = WebSocket.using[String] { implicit request =>
  SecureSocial.currentUser(request) match {
    case Some(userid) => ... success
    case None => ... failed
  }
}

【讨论】：