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

Question

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 filename

Answer 1

The Travis answer is correct but I have tweaked a little to get a faster version of his code:

val Buffer = 2 * 1024

def zip(out: String, files: Iterable[String], retainPathInfo: Boolean = true) = {
  var data = new Array[Byte](Buffer)
  val zip = new ZipOutputStream(new FileOutputStream(out))
  files.foreach { name =>
    if (!retainPathInfo)
      zip.putNextEntry(new ZipEntry(name.splitAt(name.lastIndexOf(File.separatorChar) + 1)._2))
    else
      zip.putNextEntry(new ZipEntry(name))
    val in = new BufferedInputStream(new FileInputStream(name), Buffer)
    var b = in.read(data, 0, Buffer)
    while (b != -1) {
      zip.write(data, 0, b)
      b = in.read(data, 0, Buffer)
    }
    in.close()
    zip.closeEntry()
  }
  zip.close()
}