How to list all files from resources folder with scala

Question

Assume the following structure in your recources folder:

resources
├─spec_A
| ├─AA
| | ├─file-aev
| | ├─file-oxa
| | ├─…
| | └─file-stl
| ├─BB
| | ├─file-hio
| | ├─file-nht
| | ├─…
| | └─file-22an
| └─…
├─spec_B
| ├─AA
| | ├─file-aev
| | ├─file-oxa
| | ├─…
| | └─file-stl
| ├─BB
| | ├─file-hio
| | ├─file-nht
| | ├─…
| | └─file-22an
| └─…
└─…

The task is to read all files for a given specification spec_X one subfolder by one. For obvious reasons we do not want to have the exact names as string literals to open with Source.fromResource("spec_A/AA/…") for hundreds of files in the code.

Additionally, this solution should of course run inside the development environment, i.e. without being packaged into a jar.

Answer 1

Thanks to @TrebledJ ’s answer, this could be minimized to the following:

class ConfigFiles (val basePath String) {
  lazy val jarFileSystem: FileSystem = FileSystems.newFileSystem(getClass.getResource(basePath).toURI, Map[String, String]().asJava);

  def listPathsFromResource(folder: String): List[Path] = {
    Files.list(getPathForResource(folder))
      .filter(p ⇒ Files.isRegularFile(p, Array[LinkOption](): _*))
      .sorted.toList.asScala.toList // from Stream to java List to Scala Buffer to scala List
  }

  private def getPathForResource(filename: String) = {
    val url = classOf[ConfigFiles].getResource(basePath + "/" + filename)
    if ("file" == url.getProtocol) Paths.get(url.toURI)
    else jarFileSystem.getPath(basePath, filename)
  }
}

special attention was necessary for the empty setting maps.

checking for the URL protocol seems inevitable. Git updated, PUll requests welcome: https://github.com/kurellajunior/list-files-from-resource-directory