需要对象：play.api.mvc.Result 错误

Question

我想用这个

来保护我的应用程序

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"""")   } }

但是在 getOrElse 部分，我收到以下错误：

type mismatch; found : Object required: play.api.mvc.Result

怎么了？

Answer 1

问题是 action(request) 返回的是 Future[Result] 而不仅仅是 Result。因此 getOrElse 之前的整个表达式的类型是 Option[Future[Result]] 并且它等待 Future[Result] 作为 getOrElse.

的参数

所以一开始你可以把你的 Unauthorized.withHeaders 包装成 Future.successfull( ... )

接下来是整个表达式的结果是 Future[Result]，它不是 Action.apply(bodyParser)( ... ) 的正确参数类型，但有 Action.async 方法可以处理它。

所以具有最小类型更正\重构的整个更正块是

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""""))
    }
  }

进一步的糖注入可能会产生更具可读性的版本：

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，<- 左侧的解包转换为 filter match

需要对象：play.api.mvc.Result 错误

Object required: play.api.mvc.Result error

scala

playframework-2.0