Scala: Working with resources, folders and files

Let’s briefly discuss how to deal with a resources folder and directories in Scala project. It’s pretty frequent case in a programming, when you need to interact with file system, Scala isn’t an exception. So how it can be done on practice?

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I’m going to demonstrate a short example on a real Scala project with a such structure:

As you see it has the resources folder with files and directories inside of it.

Access the resources folder

In order to get a path of files from the resources folder, I need to use following code:

object Demo {

  def main(args: Array[String]): Unit = {
    val resourcesPath = getClass.getResource("/json-sample.js")
    println(resourcesPath.getPath)

  }
}

An output of the code below, is something like this:

/Users/Alex/IdeaProjects/DirectoryFiles/out/production/DirectoryFiles/json-sample.js

Read file line by line

In some sense a pure path to a file is useless. Let’s try to read the file from resources line by line. Scala can do it well:

import scala.io.Source

object Demo {

  def main(args: Array[String]): Unit = {

    val fileStream = getClass.getResourceAsStream("/json-sample.js")
    val lines = Source.fromInputStream(fileStream).getLines
    lines.foreach(line => println(line))

  }

}

The output is

{
    "name": "Alex",
    "age": 26
}

List files in directory

The final important and popular task is to list files from a directory.

import java.io.File

object Demo {

  def main(args: Array[String]): Unit = {

    val path = getClass.getResource("/folder")
    val folder = new File(path.getPath)
    if (folder.exists && folder.isDirectory)
      folder.listFiles
        .toList
        .foreach(file => println(file.getName))

  }
}

Looks simple and efficient as well.

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