如何使用scala将多个文件存档到.zip文件中?

时间:2012-04-03 00:38:33

标签: scala zipfile

有人可以发布一个简单的代码片段吗?

文件是文本文件,因此压缩会很好,而不仅仅是归档文件。

我将文件名存储在iterable中。

6 个答案:

答案 0 :(得分:20)

目前没有任何方法可以从标准Scala库中执行此类操作,但使用java.util.zip非常简单:

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()
}

我在这里专注于简单而不是效率(没有错误检查,一次读取和写入一个字节并不理想),但它可以工作,并且可以很容易地进行改进。

答案 1 :(得分:7)

我最近也不得不使用zip文件,并发现了这个非常好的实用程序:https://github.com/zeroturnaround/zt-zip

以下是压缩目录中所有文件的示例:

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

非常方便。

答案 2 :(得分:3)

如果你喜欢功能,这是一个更多的scala风格:

  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()
    }
  }

并且不要忘记导入:

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

答案 3 :(得分:3)

特拉维斯的答案是正确的,但我已经调整了一点,以获得更快的代码版本:

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()
}

答案 4 :(得分:1)

使用NIO2的位修改(更短)版本:

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()
}

答案 5 :(得分:0)

正如Gabriele Petronella所建议的那样,此外需要在pom.xml中添加以下maven依赖项以及以下导入

*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>*