Question

我想用这个

保护我的应用程序

def Secured[A](username: String, password: String)(action: Action[A]) = Action(action.parser) { 
 request =>   request.headers.get("Authorization").flatMap { authorization =>
    authorization.split(" ").drop(1).headOption.filter { 
       encoded => new String(org.apache.commons.codec.binary.Base64.decodeBase64(encoded.getBytes)).split(":").toList match {
        case u :: p :: Nil if u == username && password == p => true
        case _ => false
      }
    }.map(_ => action(request))   }.getOrElse {
    Unauthorized.withHeaders("WWW-Authenticate" -> """Basic realm="Secured"""")   } }

但在getOrElse部分，我收到以下错误：

类型不匹配; found：需要对象：play.api.mvc.Result

有什么问题？

Answer 1

此处action(request)正在返回Future[Result]，而不仅仅是Result。因此，getOrElse之前的整个表达式类型为Option[Future[Result]]，它等待Future[Result]作为getOrElse的参数。

首先，您可以将Unauthorized.withHeaders内容包装到Future.successfull( ... )

中

接下来的事情是整个表达式的结果是Future[Result]并且它不是Action.apply(bodyParser)( ... )的正确参数类型，但是有Action.async方法可以处理它。

所以整个修正的块具有最小类型修正\重构

import org.apache.commons.codec.binary.Base64

def secured[A](username: String, password: String)(action: Action[A]) =
  Action.async(action.parser) {
    request => request.headers.get("Authorization").flatMap { authorization =>
      authorization.split(" ").drop(1).headOption.filter { encoded =>
        val bytes = Base64.decodeBase64(encoded.getBytes)
        new String(bytes).split(":").toList match {
          case u :: p :: Nil if u == username && password == p => true
          case _ => false
        }
      }.map(_ => action(request))
    }.getOrElse {
      Future.successful(
        Unauthorized.withHeaders(
          "WWW-Authenticate" -> """Basic realm="Secured""""))
    }
  }

进一步注射糖可能会导致更具可读性的版本：

import org.apache.commons.codec.binary.Base64

def secured[A](username: String, password: String)(action: Action[A]) =
  Action.async(action.parser) { request =>
    val result = for {
      authorization <- request.headers.get("Authorization")
      Array(_, encoded, _*) <- Some(authorization.split(" "))
      bytes = Base64.decodeBase64(encoded.getBytes)
      credentials = new String(bytes).split(":").toList
      List(`username`, `password`) <- Some(credentials)
    } yield action(request)

    result.getOrElse {
      Future.successful(
        Unauthorized.withHeaders(
          "WWW-Authenticate" -> """Basic realm="Secured""""))
    }
  }

请注意，<-左侧的解包已转换为filter match

需要的对象：play.api.mvc.Result错误

1 个答案: