How to get filename of xlsx file with apache poi XSSF?

Question

How to get filename of xlsx file with apache poi XSSF?

case class XlsxSplitter(path: InputStream){

  lazy val spreadSheet=load(path)

  def load(path: InputStream):SpreadSheet={
    val wb = new XSSFWorkbook(path)
    .........
  }
}

I could extract it from the path, but I would like to make my case class as generic as possible.

Answer 1

If you're able to change the path attribute to a attribute of type File instead of InputStream, you can get the filename from the file itself by file.getName();

Otherwise I think you have no other choice than extracting it by yourself.