在Scala中将Spark SQL函数名作为参数传递

import org.apache.spark.sql.{SparkSession, functions => F}

object DynamicFunctionExample {
  def main(args: Array[String]): Unit = {
    val spark = SparkSession.builder()
      .appName("Dynamic Function Example")
      .master("local[*]")
      .getOrCreate()

    import spark.implicits._

    // 创建一个示例DataFrame
    val data = Seq(
      ("Alice", 29),
      ("Bob", 31),
      ("Catherine", 25)
    ).toDF("name", "age")

    // 动态函数名
    val functionName = "avg"

    // 根据函数名动态调用相应的Spark SQL函数
    val result = functionName match {
      case "count" => data.select(F.count($"name")).head().getLong(0)
      case "sum" => data.select(F.sum($"age")).head().getLong(0)
      case "avg" => data.select(F.avg($"age")).head().getDouble(0)
      case _ => throw new IllegalArgumentException(s"Unsupported function: $functionName")
    }

    println(s"Result of $functionName: $result")

    spark.stop()
  }
}