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

This is a little bit more scala style in case you like functional:

  def compress(zipFilepath: String, files: List[File]) {
    def readByte(bufferedReader: BufferedReader): Stream[Int] = {
      bufferedReader.read() #:: readByte(bufferedReader)
    }
    val zip = new ZipOutputStream(new FileOutputStream(zipFilepath))
    try {
      for (file <- files) {
        //add zip entry to output stream
        zip.putNextEntry(new ZipEntry(file.getName))

        val in = Source.fromFile(file.getCanonicalPath).bufferedReader()
        try {
          readByte(in).takeWhile(_ > -1).toList.foreach(zip.write(_))
        }
        finally {
          in.close()
        }

        zip.closeEntry()
      }
    }
    finally {
      zip.close()
    }
  }

and don't forget the imports:

import java.io.{BufferedReader, FileOutputStream, File}
import java.util.zip.{ZipEntry, ZipOutputStream}
import io.Source