如何使用 Scala 将多个文件归档到 .zip 文件中? [英] How do I archive multiple files into a .zip file using scala?

问题描述

有人可以发布一个简单的代码片段吗?

Could anyone post a simple snippet that does this?

文件是文本文件，因此压缩比归档文件更好.

Files are text files, so compression would be nice rather than just archive the files.

我将文件名存储在可迭代对象中.

I have the filenames stored in an iterable.

推荐答案

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

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()
}

我在这里关注的是简单性而不是效率(没有错误检查并且一次读写一个字节并不理想)，但它确实有效，并且可以很容易地改进.

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.

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