使用databricks在Spark(scala)中生成带有属性和值的XML

import org.apache.spark.sql.{Row, SparkSession}
import com.databricks.spark.xml._

val spark = SparkSession.builder()
  .appName("XML Generation")
  .getOrCreate()

val xmlData = Seq(
  Row("<person id=\"1\"><name>John</name><age>30</age></person>"),
  Row("<person id=\"2\"><name>Jane</name><age>25</age></person>")
)
val schema = new StructType().add("xml", StringType)

4. 将数据转换为DataFrame：
```scala
val df = spark.createDataFrame(spark.sparkContext.parallelize(xmlData), schema)

val xmlDF = df.select(to_xml(struct($"xml")).alias("xml"))

xmlDF.write
  .format("xml")
  .option("rootTag", "people")
  .option("rowTag", "person")
  .save("path/to/save/xml")