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我在 ListBuffer 上创建了循环:

val listOfNamesOfMissingFiles = new ListBuffer[String]()
controlFileContent.entries.foreach { entry =>
      val fileName = entry("File_Name")
      val sha256 = entry("File_SHA_256")
      val file = new File(folder, fileName)
      if (!file.exists()) {
        logger.error(s"Control file validation failed: file does not exist in folder: [$fileName].")
        listOfNamesOfMissingFiles += fileName
      }

我想将我在 listOfNamesOfMissingFiles 中获得的内容分配给一个属性,并将其作为事件发送到 kibana 门户。如果我的 listOfNamesOfMissingFiles == ListBuffer("hhhh","uuuu"). 如果我会做 listOfNamesOfMissingFiles.toString 我会得到“ListBuffer(“hhhh”,“uuuu”)”。如何从输出中删除 ListBuffer?

4

1 回答 1

5

我认为你想要实现的只是加入一个字符串列表。Scala提供了mkString方法:

scala> val names = ListBuffer("hhhh", "uuuu")
names: scala.collection.mutable.ListBuffer[String] = ListBuffer(hhhh, uuuu)

scala> names.mkString
res0: String = hhhhuuuu

scala> names.mkString(", ")
res1: String = hhhh, uuuu

scala> names.mkString("(", ", ", ")")
res2: String = (hhhh, uuuu)
于 2018-12-23T14:24:22.000 回答