如何从string scala中提取文件扩展名

val filePath = "path/to/file.txt"
val fileNameAndExtension = filePath.split("\\.")

val fileExtension = fileNameAndExtension.last

val lowercaseExtension = fileExtension.toLowerCase

val filePath = "path/to/file.txt"
val fileNameAndExtension = filePath.split("\\.")
val fileExtension = fileNameAndExtension.last
val lowercaseExtension = fileExtension.toLowerCase

println(lowercaseExtension)