Question

我有以下代码向HTTP服务请求获取资源。可能是服务不可用或资源不可用。所以我需要处理这样的错误情况。服务器实际上抛出了我必须在我的Scala客户端中处理的异常。以下是我到目前为止所做的事情：

  def getData(resource: String, requestParam: Option[Seq[(String, String)]]): Try[Elem] = {
    Try {
      val wc = WebClient.create(address(scheme, host, port) + "/" + resource, un, pw, null)

      // Load the trust store to the HTTPConduit
      if ("https" == scheme) {
        val conduit = WebClient.getConfig(wc).getConduit.asInstanceOf[HTTPConduit]
        loadTrustStore(conduit)
      }

      requestParam match {
        case Some(reqParams) => reqParams.foreach((param: (String, String)) => {
          wc.query(param._1, param._2)
        })
        case None => wc
      }
      XML.loadString("""<?xml version="1.0" encoding="UTF-8"?>""" + wc.get(classOf[String]))
    }
  }

现在，此方法的调用者如下所示：

def prepareDomainObject(): MyDomainObject = {
  getData(resource, params) match {
  case Success(elem) => getDomainFromElem(elem)
  case Failure(_) => ?? What do I do here? I actually want to get the message in the Throwable and return another type. How do I get around this?
}

如何处理匹配条件，以便在出现错误时返回其他类型？

Answer 1

您有两种选择。要么prepareDomainObject也要返回Try：

def prepareDomainObject(): Try[MyDomainObject] = {
  getData(resource, params) map getDomainFromElem
}

或者在错误时抛出异常：

def prepareDomainObject(): MyDomainObject = {
  getDomainFromElem(getData(resource, params).get)
}

Scala异常处理 - 需要建议

1 个答案: