如何使用scala解压zip文件?

Basically I need to unzip a .zip file which contains a folder called modeled which in turn contains a number of excel files.

基本上，我需要解压缩一个.zip文件，其中包含一个名为model的文件夹，该文件夹中包含许多excel文件。

I have had some luck in finding code that was already written (ZipArchive) which is meant to unzip the zip file, but I cannot figure out why it throws an error message when I use it. The code for ZipArchive and the error message are listed below:

我很幸运地找到了已经编写的(ZipArchive)代码，该代码是用来解压zip文件的，但是我无法理解为什么在使用它时它会抛出错误消息。ZipArchive的代码和错误消息如下:

import java.io.{OutputStream, InputStream, File, FileOutputStream}
import java.util.zip.{ZipEntry, ZipFile}
import scala.collection.JavaConversions._

object ZipArchive {

  val BUFSIZE = 4096
  val buffer = new Array[Byte](BUFSIZE)

  def unZip(source: String, targetFolder: String) = {
    val zipFile = new ZipFile(source)

    unzipAllFile(zipFile.entries.toList, getZipEntryInputStream(zipFile)_, new File(targetFolder))
  }

  def getZipEntryInputStream(zipFile: ZipFile)(entry: ZipEntry) = zipFile.getInputStream(entry)

  def unzipAllFile(entryList: List[ZipEntry], inputGetter: (ZipEntry) => InputStream, targetFolder: File): Boolean = {

    entryList match {
      case entry :: entries =>

        if (entry.isDirectory)
          new File(targetFolder, entry.getName).mkdirs
        else
          saveFile(inputGetter(entry), new FileOutputStream(new File(targetFolder, entry.getName)))

        unzipAllFile(entries, inputGetter, targetFolder)
      case _ =>
        true
    }
  }

  def saveFile(fis: InputStream, fos: OutputStream) = {
    writeToFile(bufferReader(fis)_, fos)
    fis.close
    fos.close
  }

  def bufferReader(fis: InputStream)(buffer: Array[Byte]) = (fis.read(buffer), buffer)

  def writeToFile(reader: (Array[Byte]) => Tuple2[Int, Array[Byte]], fos: OutputStream): Boolean = {
    val (length, data) = reader(buffer)
    if (length >= 0) {
      fos.write(data, 0, length)
      writeToFile(reader, fos)
    } else
      true
  }
}

Error Message:

错误信息:

java.io.FileNotFoundException: src/test/resources/oepTemp/modeled/EQ_US_2_NULL_('CA')_ALL_ELT_IL_EQ_US.xlsx (No such file or directory), took 6.406 sec
[error]     at java.io.FileOutputStream.open(Native Method)
[error]     at java.io.FileOutputStream.<init>(FileOutputStream.java:221)
[error]     at java.io.FileOutputStream.<init>(FileOutputStream.java:171)
[error]     at com.contract.testing.ZipArchive$.unzipAllFile(ZipArchive.scala:28)
[error]     at com.contract.testing.ZipArchive$.unZip(ZipArchive.scala:15)
[error]     at com.contract.testing.OepStepDefinitions$$anonfun$1.apply$mcZ$sp(OepStepDefinitions.scala:175)
[error]     at com.contract.testing.OepStepDefinitions$$anonfun$1.apply(OepStepDefinitions.scala:150)
[error]     at com.contract.testing.OepStepDefinitions$$anonfun$1.apply(OepStepDefinitions.scala:150)
[error]     at cucumber.api.scala.ScalaDsl$StepBody$$anonfun$apply$1.applyOrElse(ScalaDsl.scala:61)
[error]     at cucumber.api.scala.ScalaDsl$StepBody$$anonfun$apply$1.applyOrElse(ScalaDsl.scala:61)
[error]     at scala.runtime.AbstractPartialFunction.apply(AbstractPartialFunction.scala:36)
[error]     at cucumber.runtime.scala.ScalaStepDefinition.execute(ScalaStepDefinition.scala:71)
[error]     at cucumber.runtime.StepDefinitionMatch.runStep(StepDefinitionMatch.java:37)
[error]     at cucumber.runtime.Runtime.runStep(Runtime.java:298)
[error]     at cucumber.runtime.model.StepContainer.runStep(StepContainer.java:44)
[error]     at cucumber.runtime.model.StepContainer.runSteps(StepContainer.java:39)
[error]     at cucumber.runtime.model.CucumberScenario.run(CucumberScenario.java:48)
[error]     at cucumber.runtime.junit.ExecutionUnitRunner.run(ExecutionUnitRunner.java:91)
[error]     at cucumber.runtime.junit.FeatureRunner.runChild(FeatureRunner.java:63)
[error]     at cucumber.runtime.junit.FeatureRunner.runChild(FeatureRunner.java:18)
[error]     ...

So based on the error message it looks like it's trying to find the exported excel file? This part completely throws me off. Any help would be greatly appreciated. I've added below how I'm calling the method, perhaps I'm doing something silly. Also I'm up for using a different way to extract my zip file if you can recommend one.

基于错误消息，它看起来像是在寻找导出的excel文件?这部分完全把我搞砸了。任何帮助都将不胜感激。我在下面添加了如何调用这个方法，也许我在做一些愚蠢的事情。如果你能推荐的话，我也很乐意用一种不同的方法提取我的zip文件。

val tempDirectoryDir = "src/test/resources/oepTemp/"
ZipArchive.unZip(tempDirectoryDir + "Sub Region Input - Output.zip", tempDirectoryDir)

4 个解决方案

#1

Well since were are using some utilities from java, here is a version basen on this, translated to scala, maybe this should be more functional, but it is useful

既然我们使用了java的一些实用工具，这里有一个关于这个的版本basen，翻译成scala，也许这应该更实用一些，但是它很有用

package zip

import java.io.{ IOException, FileOutputStream, FileInputStream, File }
import java.util.zip.{ ZipEntry, ZipInputStream }

/**
 * Created by anquegi on 04/06/15.
 */
object Unzip extends App {

  val INPUT_ZIP_FILE: String = "src/main/resources/my-zip.zip";
  val OUTPUT_FOLDER: String = "src/main/resources/my-zip";

  def unZipIt(zipFile: String, outputFolder: String): Unit = {

    val buffer = new Array[Byte](1024)

    try {

      //output directory
      val folder = new File(OUTPUT_FOLDER);
      if (!folder.exists()) {
        folder.mkdir();
      }

      //zip file content
      val zis: ZipInputStream = new ZipInputStream(new FileInputStream(zipFile));
      //get the zipped file list entry
      var ze: ZipEntry = zis.getNextEntry();

      while (ze != null) {

        val fileName = ze.getName();
        val newFile = new File(outputFolder + File.separator + fileName);

        System.out.println("file unzip : " + newFile.getAbsoluteFile());

        //create folders
        new File(newFile.getParent()).mkdirs();

        val fos = new FileOutputStream(newFile);

        var len: Int = zis.read(buffer);

        while (len > 0) {

          fos.write(buffer, 0, len)
          len = zis.read(buffer)
        }

        fos.close()
        ze = zis.getNextEntry()
      }

      zis.closeEntry()
      zis.close()

    } catch {
      case e: IOException => println("exception caught: " + e.getMessage)
    }

  }

  Unzip.unZipIt(INPUT_ZIP_FILE, OUTPUT_FOLDER)

}

#2

Here's a more functional and precise way doing this

这里有一种更实用、更精确的方法

import java.io.{FileInputStream, FileOutputStream}
import java.util.zip.ZipInputStream
val fis = new FileInputStream("htl.zip")
val zis = new ZipInputStream(fis)
Stream.continually(zis.getNextEntry).takeWhile(_ != null).foreach{ file =>
    val fout = new FileOutputStream(file.getName)
    val buffer = new Array[Byte](1024)
    Stream.continually(zis.read(buffer)).takeWhile(_ != -1).foreach(fout.write(buffer, 0, _))
}

#3

Trying to work with Tian-Liang's solution, I realized that its not working for zips with a directory structure. So I adopted it this way:

在尝试使用天亮的解决方案时，我意识到它不适用于目录结构的zips。所以我采用了这种方式:

  import java.io.{FileOutputStream, InputStream}
  import java.nio.file.Path
  import java.util.zip.ZipInputStream

  def unzip(zipFile: InputStream, destination: Path): Unit = {
    val zis = new ZipInputStream(zipFile)

    Stream.continually(zis.getNextEntry).takeWhile(_ != null).foreach { file =>
      if (!file.isDirectory) {
        val outPath = destination.resolve(file.getName)
        val outPathParent = outPath.getParent
        if (!outPathParent.toFile.exists()) {
          outPathParent.toFile.mkdirs()
        }

        val outFile = outPath.toFile
        val out = new FileOutputStream(outFile)
        val buffer = new Array[Byte](4096)
        Stream.continually(zis.read(buffer)).takeWhile(_ != -1).foreach(out.write(buffer, 0, _))
      }
    }
  }

#4

import java.io.FileInputStream
import java.io.InputStream
import java.util.zip.ZipEntry
import java.util.zip.ZipInputStream
import scala.language.reflectiveCalls
import scala.util.Try

import org.apache.commons.io.IOUtils

def using[T <: { def close() }, U](resource: T)(block: T => U): U = {
  try {
    block(resource)
  } finally {
    if (resource != null) {
        resource.close()
    }
  }
}

def processZipFile(zipFile: ZipFile)(doStuff: ZipEntry => Unit) {
  using(new ZipInputStream(new FileInputStream(zipFile))) { zipInputStream =>
    val entries = Stream.continually(Try(zipInputStream.getNextEntry()).getOrElse(null))
        .takeWhile(_ != null) // while not EOF and not corrupted
        .foreach(doStuff)
        .force
  }
}