Question

我没有找到一种方法让身体进入Play 2.5.x中的Filter。我想创建一个＆＃34; BadRequestLogFilter＆＃34;，如果我的appliation返回状态代码400-500，则应该记录请求和结果

在Play 2.4.x中，我使用了Iteratees并且它有效。我无法将这段代码迁移到Play 2.5.x.有人能给我一个暗示吗？也许在过滤器中获取物体的孔方法是个坏主意？

这是我的旧版（在2.4.x中正常工作）Filter for Play 2.4.x

class BadRequestLogFilter @Inject() (implicit val mat: Materializer, ec: ExecutionContext) extends Filter {
  val logger = Logger("bad_status").underlyingLogger

  override def apply(next: (RequestHeader) => Future[Result])(request: RequestHeader): Future[Result] = {
    val resultFuture = next(request)
    resultFuture.foreach(result => {

      val status = result.header.status
      if (status < 200 || status >= 400) {
        val c = Try(request.tags(Router.Tags.RouteController))
        val a = Try(request.tags(Router.Tags.RouteActionMethod))

        val body = result.body.run(Iteratee.fold(Array.empty[Byte]) { (memo, nextChunk) => memo ++ nextChunk })
        val futResponse = body.map(bytes => new String(bytes))
        futResponse.map { response =>

          val m = Map("method" -> request.method,
            "uri" -> request.uri,
            "status" -> status,
            "response" -> response,
            "request" -> request,
            "controller" -> c.getOrElse("empty"),
            "actionMethod" -> a.getOrElse("empty"))

          val msg = m.map { case (k, v) => s"$k=$v" }.mkString(", ")
          logger.info(appendEntries(m), msg)
        }
      }
    })
    resultFuture
  }
}

我想我这里只需要一个有效的替代品：

val body = result.body.run(Iteratee.fold(Array.empty[Byte]) { (memo, nextChunk) => memo ++ nextChunk })

Answer 1

Play 2.5.x中的结果正文属于HttpEntity

因此，一旦获得结果，您就可以获得身体然后实现它：

val body = result.body.consumeData(mat)

此处mat是您拥有的implicit Materializer。这将返回一个Future[ByteString]然后您可以解码以获得字符串表示（为简单起见，我在此处省略了将来的处理）：

val bodyAsString = body.decodeString("UTF-8")
logger.info(bodyAsString)

Scala Play 2.5.0 2.5.1可以访问结果正文进行过滤

1 个答案: