在 Play 框架中处理异常

Question

我想对异常处理机制提出一些建议。我目前终于有了传统的 try catch，但我知道它在风格上不起作用。那么处理异常的最佳功能方式是什么。我尝试查看 Scala 手臂，但我想它最终只是 try catch 的功能包装器！建议？这是我的 Play 控制器，我想在其中处理抛出的异常并将纯字符串发送回客户端！

def testmethod = Action(parse.maxLength(10 * 1024 * 1024, parser = parse.anyContent)) { implicit request =>
    request.body match {
      case Left(_) => EntityTooLarge
      case Right(body) => {
        println(request.headers.toSimpleMap)
        val js = // Get the value from the request!!!

        val resultJsonfut = scala.concurrent.Future { longRunningProcess(js) }

        Async {
          if(request.headers.toSimpleMap.exists(_ == (ACCEPT_ENCODING, "gzip"))) {
            resultJsonfut.map(s => {
              val bytePayload = getBytePayload(s) // How to handle exceptions thrown by getBytePayLoad????
              Ok(bytePayload)
            })
          } else {
            resultJsonfut.map(s => Ok(s))
          }
        }
      }
    }
  }

Answer 1

你可以创建一个方法来处理你的动作中的异常，这个方法看起来像这样：

def withExceptionHandling(f: => Result)(implicit request: Request[AnyContent]): Result = 
    try{ f }catch{ case e: Exception => InternalServerError("Something bad happened")}

然后你会像这样使用它：

def testmethod = Action(parse.maxLength(10 * 1024 * 1024, parser = parse.anyContent)) { implicit request => 
  withExceptionHandling {
    request.body match {
      case Left(_) => EntityTooLarge
      case Right(body) => {
        println(request.headers.toSimpleMap)
        val js = // Get the value from the request!!!

        val resultJsonfut = scala.concurrent.Future { longRunningProcess(js) }

        Async {
          if(request.headers.toSimpleMap.exists(_ == (ACCEPT_ENCODING, "gzip"))) {
            resultJsonfut.map(s => {
              val bytePayload = getBytePayload(s) // How to handle exceptions thrown by getBytePayLoad????
              Ok(bytePayload)
            })
          } else {
            resultJsonfut.map(s => Ok(s))
          }
        }
      }
    }
  }
}

这将阻止你明确地尝试和捕捉你的每一个动作。