如何遍历在单个scala.xml.Node中找到的子项列表

val node: scala.xml.Node = ...
val children: scala.xml.NodeSeq = node.child

children.foreach { child =>
  // 在此处处理每个子节点
}

val transformedChildren: Seq[String] = children.map { child =>
  // 在此处将每个子节点转换为所需类型
  child.text
}

val filteredChildren: scala.xml.NodeSeq = children.filter { child =>
  // 在此处过滤子节点
  child.label == "elementName"
}

val collectedChildren: Seq[String] = children.collect {
  case child: scala.xml.Text => child.text
}

val foundChild: Option[scala.xml.Node] = children.find { child =>
  // 在此处查找满足条件的子节点
  child.label == "elementName"
}

val hasChild: Boolean = children.exists { child =>
  // 在此处检查子节点是否满足条件
  child.label == "elementName"
}

val foldedValue: Int = children.foldLeft(0) { (acc, child) =>
  // 在此处将子节点折叠为单个值
  acc + child.text.toInt
}

val reducedValue: Int = children.reduceLeft { (acc, child) =>
  // 在此处将子节点折叠为单个值
  acc + child.text.toInt
}

val indexedChildren: scala.xml.NodeSeq = children.zipWithIndex.map {
  case (child, index) =>
    // 在此处处理每个子节点及其索引
    child
}