How do I archive multiple files into a .zip file using scala? - 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 filenames stored in an iterable.

Answer 1

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.

Answer 2

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.

Answer 3

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

Answer 4

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()
}

Answer 5

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()
}

Answer 6

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>*