How do I list all files in a subdirectory in scala?

Question

Is there a good \"scala-esque\" (I guess I mean functional) way of recursively listing files in a directory? What about matching a particular pattern?

For example re

Answer 1

The simplest Scala-only solution (if you don't mind requiring the Scala compiler library):

val path = scala.reflect.io.Path(dir)
scala.tools.nsc.io.Path.onlyFiles(path.walk).foreach(println)

Otherwise, @Renaud's solution is short and sweet (if you don't mind pulling in Apache Commons FileUtils):

import scala.collection.JavaConversions._  // enables foreach
import org.apache.commons.io.FileUtils
FileUtils.listFiles(dir, null, true).foreach(println)

Where dir is a java.io.File:

new File("path/to/dir")

Answer 2

Scala code typically uses Java classes for dealing with I/O, including reading directories. So you have to do something like:

import java.io.File
def recursiveListFiles(f: File): Array[File] = {
  val these = f.listFiles
  these ++ these.filter(_.isDirectory).flatMap(recursiveListFiles)
}

You could collect all the files and then filter using a regex:

myBigFileArray.filter(f => """.*\.html$""".r.findFirstIn(f.getName).isDefined)

Or you could incorporate the regex into the recursive search:

import scala.util.matching.Regex
def recursiveListFiles(f: File, r: Regex): Array[File] = {
  val these = f.listFiles
  val good = these.filter(f => r.findFirstIn(f.getName).isDefined)
  good ++ these.filter(_.isDirectory).flatMap(recursiveListFiles(_,r))
}

Answer 3

I like yura's stream solution, but it (and the others) recurses into hidden directories. We can also simplify by making use of the fact that listFiles returns null for a non-directory.

def tree(root: File, skipHidden: Boolean = false): Stream[File] = 
  if (!root.exists || (skipHidden && root.isHidden)) Stream.empty 
  else root #:: (
    root.listFiles match {
      case null => Stream.empty
      case files => files.toStream.flatMap(tree(_, skipHidden))
  })

Now we can list files

tree(new File(".")).filter(f => f.isFile && f.getName.endsWith(".html")).foreach(println)

or realise the whole stream for later processing

tree(new File("dir"), true).toArray

Answer 4

Take a look at scala.tools.nsc.io

There are some very useful utilities there including deep listing functionality on the Directory class.

If I remember correctly this was highlighted (possibly contributed) by retronym and were seen as a stopgap before io gets a fresh and more complete implementation in the standard library.

Answer 5

Scala has library 'scala.reflect.io' which considered experimental but does the work

import scala.reflect.io.Path
Path(path) walkFilter { p => 
  p.isDirectory || """a*.foo""".r.findFirstIn(p.name).isDefined
}

Answer 6

Apache Commons Io's FileUtils fits on one line, and is quite readable:

import scala.collection.JavaConversions._ // important for 'foreach'
import org.apache.commons.io.FileUtils

FileUtils.listFiles(new File("c:\temp"), Array("foo"), true).foreach{ f =>

}