如何在spark scala中为每个记录创建一个xml文件

import org.apache.spark.sql.{SparkSession, Row}
import org.apache.spark.sql.functions._
import org.apache.spark.sql.types._
import javax.xml.parsers.DocumentBuilderFactory
import javax.xml.transform.dom.DOMSource
import javax.xml.transform.stream.StreamResult
import javax.xml.transform.TransformerFactory

val spark = SparkSession.builder()
  .appName("CreateXMLFiles")
  .getOrCreate()

val schema = StructType(Seq(
  StructField("id", IntegerType, nullable = false),
  StructField("name", StringType, nullable = false),
  StructField("age", IntegerType, nullable = false)
))

val data = Seq(
  Row(1, "John", 25),
  Row(2, "Jane", 30),
  Row(3, "Bob", 35)
)

val df = spark.createDataFrame(spark.sparkContext.parallelize(data), schema)

def createXMLFile(row: Row): Unit = {
  val docBuilder = DocumentBuilderFactory.newInstance().newDocumentBuilder()
  val doc = docBuilder.newDocument()

  val rootElement = doc.createElement("record")
  doc.appendChild(rootElement)

  val idElement = doc.createElement("id")
  idElement.appendChild(doc.createTextNode(row.getInt(0).toString))
  rootElement.appendChild(idElement)

  val nameElement = doc.createElement("name")
  nameElement.appendChild(doc.createTextNode(row.getString(1)))
  rootElement.appendChild(nameElement)

  val ageElement = doc.createElement("age")
  ageElement.appendChild(doc.createTextNode(row.getInt(2).toString))
  rootElement.appendChild(ageElement)

  val transformer = TransformerFactory.newInstance().newTransformer()
  transformer.transform(new DOMSource(doc), new StreamResult(new File(s"${row.getInt(0)}.xml")))
}

df.foreach(row => createXMLFile(row))