无法解析作为字符串传递的列的列名错误- Spark Scala

val df = Seq(("Alice", 29), ("Bob", 31)).toDF("name", "age")
val result = df.select("name") // 正确

import org.apache.spark.sql.functions.col

val result = df.select(col("name")) // 正确

if (df.columns.contains("name")) {
  val result = df.select("name")
} else {
  println("列名不存在")
}

val df = Seq(("Alice", 29), ("Bob", 31)).toDF("name", "age")
val dfRenamed = df.withColumnRenamed("nme", "name") // 将错误的列名重命名为正确的列名
val result = dfRenamed.select("name")

import org.apache.spark.sql.SparkSession
import org.apache.spark.sql.functions.col

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

val data = Seq(("Alice", 29), ("Bob", 31))
val df = data.toDF("name", "age")

// 正确的列名引用
val result = df.select(col("name"))
result.show()

// 错误的列名引用
try {
  val errorResult = df.select("nme")
} catch {
  case e: Exception => println(s"错误: ${e.getMessage}")
}

spark.stop()