如何使用 scala 列出资源文件夹中的所有文件

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, ie without being packaged into a jar.

Answer 1

The only option to list files inside a resource folder I found is with nio's Filesystem concept, as this can load a jar-file as a file system. But this comes with two major downsides:

1. java.nio uses java Stream API, which I cannot collect from inside scala code: Collectors.toList() cannot be made to compile as it cannot determine the right type.
2. The filesystem needs different base paths for OS-filesystems and jar-file-based filesystems. So I need to manually differentiate between the two situations testing and jar-based running.

First lazy load the jar-filesystem if needed

  private static FileSystem jarFileSystem;

  static synchronized private FileSystem getJarFileAsFilesystem(String drg_file_root) throws URISyntaxException, IOException {
    if (jarFileSystem == null) {
      jarFileSystem = FileSystems.newFileSystem(ConfigFiles.class.getResource(drg_file_root).toURI(), Collections.emptyMap());
    }
    return jarFileSystem;
  }

next do the limbo to figure out whether we are inside the jar or not by checking the protocol of the URL and return a Path. (Protocol inside the jar file will be jar:

  static Path getPathForResource(String resourceFolder, String filename) throws IOException, URISyntaxException {
    URL url = ConfigFiles.class.getResource(resourceFolder + "/" + filename);
    return "file".equals(url.getProtocol())
           ? Paths.get(url.toURI())
           : getJarFileAsFilesystem(resourceFolder).getPath(resourceFolder, filename);
  }

And finally list and collect into a java list

  static List<Path> listPathsFromResource(String resourceFolder, String subFolder) throws IOException, URISyntaxException {
    return Files.list(getPathForResource(resourceFolder, subFolder))
      .filter(Files::isRegularFile)
      .sorted()
      .collect(toList());
  }

Only then we can go back do Scala and fetch is

class SpecReader {
  def readSpecMessage(spec: String): String = {
    List("CN", "DO", "KF")
      .flatMap(ConfigFiles.listPathsFromResource(s"/spec_$spec", _).asScala.toSeq)
      .flatMap(path ⇒ Source.fromInputStream(Files.newInputStream(path), "UTF-8").getLines())
      .reduce(_ + " " + _)
  }
}

object Main {
  def main(args: Array[String]): Unit = {
    System.out.println(new SpecReader().readSpecMessage(args.head))
  }
}

I put a running mini project to proof it here: https://github.com/kurellajunior/list-files-from-resource-directory

But of course this is far from optimal. I wanto to elmiminate the two downsides mentioned above so, that

1. scala files only
2. no extra testing code in my production library

Answer 2

Ok, after some trying and analyzing the Collector's API I was able to create a scala.List via ListBuffer collector.

class SpecReader (val spec:String) {

  private val basePath = s"/spec_$spec"
  lazy val jarFileSystem: FileSystem = FileSystems.newFileSystem(getClass.getResource(basePath).toURI, Map[String, String]().asJava);


  def readSpecMessageScala(): String = {
    List("CN", "DO", "KF")
      .flatMap(listPathsFromResource)
      .flatMap(path ⇒ Source.fromInputStream(Files.newInputStream(path), "UTF-8").getLines())
      .reduce(_ + " " + _)
  }

  val collector: Collector[_ >: Path, ListBuffer[Path], List[Path]] =  Collector.of(
    new Supplier[ListBuffer[Path]]() {
      override def get(): ListBuffer[Path] = ListBuffer[Path]()
    },
    new BiConsumer[ListBuffer[Path], Path]() {
      override def accept(t: ListBuffer[Path], u: Path): Unit = t.addOne(u)
    },
    new BinaryOperator[ListBuffer[Path]]() {
      override def apply(t: ListBuffer[Path], u: ListBuffer[Path]): ListBuffer[Path] = t.addAll(u)
    },
    new Function[ListBuffer[Path], List[Path]](){
      override def apply(v1: ListBuffer[Path]): List[Path] = v1.toList
    },
    Array[Collector.Characteristics](): _*
)

  def listPathsFromResource(folder: String): List[Path] = {
    Files.list(getPathForResource(folder))
      .filter(p ⇒ Files.isRegularFile(p, Array[LinkOption](): _*))
      .sorted.collect(collector)
  }

  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)
  }
}

object Main {
  def main(args: Array[String]): Unit = {
    System.out.println(new SpecReader(args.head).readSpecMessage())
  }
}

special attention was necessary for the empty varargs and empty setting maps.

Still the case for testing and jar operation. Git updated, PUll requests welcome: https://github.com/kurellajunior/list-files-from-resource-directory

Answer 3

Here's a function for reading all files from a resource folder. My use case is with small files. Inspired by Jan's answers, but without needing a user-defined collector or messing with Java.

// Given a folder (e.g. "A"), reads all files under the resource folder (e.g. "src/main/resources/A/**") as a Seq[String].
def readFilesFromResource(folder: String): Seq[String] = {
    // Helper for reading an individual file.
    def read(path: Path): String =
      Source.fromInputStream(Files.newInputStream(path), "UTF-8").getLines.mkString("\n")

    val uri = Main.getClass.getResource("/" + folder).toURI
    Files.list(Paths.get(uri)).collect(Collectors.toList()).asScala.map(read) // Magic!
}

^{(not catered to example in question)}

Relevant imports:

import java.nio.file._
import scala.collection.JavaConverters._ // Needed for .asScala
import java.util.stream._
import scala.io.Source