Pico Pico - 2 months ago 12x
Scala Question

How do I archive multiple files into a .zip file using scala?

Could anyone post a simple snippet that does this?

Files are text files, so compression would be nice rather than just archive the files.

I have the filenames stored in an iterable.


There's not currently any way to do this kind of thing from the standard Scala library, but it's pretty easy to use java.util.zip:

def zip(out: String, files: Iterable[String]) = {
  import java.io.{ BufferedInputStream, FileInputStream, FileOutputStream }
  import java.util.zip.{ ZipEntry, ZipOutputStream }

  val zip = new ZipOutputStream(new FileOutputStream(out))

  files.foreach { name =>
    zip.putNextEntry(new ZipEntry(name))
    val in = new BufferedInputStream(new FileInputStream(name))
    var b = in.read()
    while (b > -1) {
      b = in.read()

I'm focusing on simplicity instead of efficiency here (no error checking and reading and writing one byte at a time isn't ideal), but it works, and can very easily be improved.