scala - 如何在不显式映射结果的情况下转发 WSResponse

Question

所以我有以下操作，我想找到一种方法来直接返回响应，而不必将其映射到每个可能的状态代码的结果，即跳过 if-else 部分。

def testAction = Action { implicit requestIn => {

   val requestOut : WSRequest = WS.url("test-domain-name:9998")

   val queryString = requestIn.queryString.map { case (k,v) => k -> v.mkString }

   val futureResponse : Future[WSResponse] = requestOut.withQueryString(queryString.toList: _*).get()

   val response = Await.result(requestOut.withQueryString(queryString.toList: _*).get(), 5 seconds)

   if(response.status == 200) {
      Ok(response.xml)
   } else {
      BadRequest(response.body)
   }

}

Answer 1

你不应该等待结果。 Play 框架支持异步操作:

def testAction = Action.async { implicit requestIn =>

    val requestOut: WSRequest = WS.url("test-domain-name:9998")

    val queryString = requestIn.queryString.map { case (k, v) => k -> v.mkString }

    val futureResponse: Future[WSResponse] = requestOut.withQueryString(queryString.toList: _*).get()

    futureResponse

}

和隐式转换:

implicit def Response2Result(response: Future[WSResponse]): Future[Result] = {
  response map {
    response =>
      val headers = response.allHeaders map {
        h => (h._1, h._2.head)
      }
      Result(ResponseHeader(response.status, headers), Enumerator(response.body.getBytes))
  }
}

见 github issue .