CODEWITHSUNDEEP
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Pico
Pico

Reputation: 603

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.

Upvotes: 19

Views: 18071

Answers (7)

gekomad
gekomad

Reputation: 565

you can use scala-compress https://github.com/gekomad/scala-compress

doc: https://gekomad.github.io/scala-compress/docs/zip/

Upvotes: 0

Dinesh Shinkar
Dinesh Shinkar

Reputation: 129

As suggested by Gabriele Petronella, in addition, you need to add below Maven dependency in pom.xml, as well as below imports:

import org.zeroturnaround.zip.ZipUtil
import java.io.File

<dependency>
    <groupId>org.zeroturnaround</groupId>
    <artifactId>zt-zip</artifactId>
    <version>1.13</version>
    <type>jar</type>
</dependency>*

Upvotes: 1

George Pligoropoulos
George Pligoropoulos

Reputation: 2939

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

Upvotes: 6

mnp
mnp

Reputation: 3302

A bit modified (shorter) version using NIO2:

private def zip(out: Path, files: Iterable[Path]) = {
  val zip = new ZipOutputStream(Files.newOutputStream(out))

  files.foreach { file =>
    zip.putNextEntry(new ZipEntry(file.toString))
    Files.copy(file, zip)
    zip.closeEntry()
  }
  zip.close()
}

Upvotes: 3

Gabriele Petronella
Gabriele Petronella

Reputation: 108101

I recently had to work with zip files too and found this very nice utility: https://github.com/zeroturnaround/zt-zip

Here's an example of zipping all files inside a directory:

import org.zeroturnaround.zip.ZipUtil
ZipUtil.pack(new File("/tmp/demo"), new File("/tmp/demo.zip"))

Very convenient.

Upvotes: 11

camikiller
camikiller

Reputation: 1870

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()
}

Upvotes: 3

Travis Brown
Travis Brown

Reputation: 139038

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) {
      zip.write(b)
      b = in.read()
    }
    in.close()
    zip.closeEntry()
  }
  zip.close()
}

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.

Upvotes: 28

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