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

问题

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.

回答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.

回答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.

回答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

回答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()
}

回答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()
}

回答6:

As suggested by Gabriele Petronella, in addition need to add below maven dependency in pom.xml as well and 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>*

来源：https://stackoverflow.com/questions/9985684/how-do-i-archive-multiple-files-into-a-zip-file-using-scala

标签

scala

zipfile