Get DataStream File Name Using Flink

Question

Get DataStream File Name Using Flink

I have a streaming process with flink working with csv files in one path. I want to know the file name for each processed file.

I am currently using this function to read csv files in a path (dataPath).

val recs:DataStream[CallCenterEvent] = env
          .readFile[CallCenterEvent](
          CsvReader.getReaderFormat[CallCenterEvent](dataPath, c._2),
          dataPath,
          FileProcessingMode.PROCESS_CONTINUOUSLY,
          c._2.fileInterval)
          .uid("source-%s-%s".format(systemConfig.name, c._1))
          .name("%s records reading".format(c._1))

And using this function to get TupleCsvInputFormat.

def getReaderFormat[T <: Product : ClassTag : TypeInformation](dataPath:String, conf:URMConfiguration): TupleCsvInputFormat[T] = {
  val typeInfo = implicitly[TypeInformation[T]]
  val format: TupleCsvInputFormat[T] = new TupleCsvInputFormat[T](new Path(dataPath), typeInfo.asInstanceOf[CaseClassTypeInfo[T]])
  if (conf.quoteCharacter != null && !conf.quoteCharacter.equals(""))
    format.enableQuotedStringParsing(conf.quoteCharacter.charAt(0))
  format.setFieldDelimiter(conf.fieldDelimiter)
  format.setSkipFirstLineAsHeader(conf.ignoreFirstLine)
  format.setLenient(true)

  return format
}