How do I get the file/folder path using Apache Commons net FTPS Client

Question

I am using Apache commons ftps client to connect to an ftps server. I have the remote file path which is a directory. This directory contains a tree of sub-directories and files. I want to get the path for each file or folder. Is there any way I can get this property? Or if there is any way I could get the parent folder path, I could concatenate the file name to it.

Answer 1

I am currently using below function to get path and size of all files under a certain directory. It gets all the file in current directory and check if it is a directory or file. If it is a directory call recursively until end of the tree, if it is a file save the path and size. You may not need these "map" things you can edit according to your needs.

Usage:

getServerFiles(ftp,"") // start from root

or

getServerFiles(ftp,"directory_name") // start from given directory

Implementation:

def getServerFiles(ftp: FTPClient, dir: String): Map[String, Double] = {
    var fileList = Array[FTPFile]()
    var base = ""
    if (dir == "") {
      fileList = ftp.listFiles
    } else {
      base = dir + "/"
      fileList = ftp.listFiles(dir)
    }
    fileList.flatMap {
      x => if (x.isDirectory) {
        getServerFiles(ftp, base + x.getName)
      } else {
        Map[String, Double](base + x.getName -> x.getSize)
      }
    }.toMap
  }