如何在Scala中通过Play Framework 2.5流压缩文件（动态）？

Question

我正在研究这个实现:https://gist.github.com/kirked/03c7f111de0e9a1f74377bf95d3f0f60，但不知道如何使用它。

 import java.io.{BufferedOutputStream, ByteArrayInputStream, ByteArrayOutputStream}
import java.util.zip.{ZipEntry, ZipOutputStream}
import akka.stream.scaladsl.{StreamConverters}
import org.apache.commons.io.FileUtils
import play.api.mvc.{Action, Controller}

class HomeController extends Controller {
  def single() = Action {
                         Ok.sendFile(
                           content = new java.io.File("C:\\Users\\a.csv"),
                           fileName = _ => "a.csv"
                         )
                       }

  def zip() = Action {
                     Ok.chunked(StreamConverters.fromInputStream(fileByteData)).withHeaders(
                       CONTENT_TYPE -> "application/zip",
                       CONTENT_DISPOSITION -> s"attachment; filename = test.zip"
                     )
                   }

  def fileByteData(): ByteArrayInputStream = {
    val fileList = List(
      new java.io.File("C:\\Users\\a.csv"),
      new java.io.File("C:\\Users\\b.csv")
    )

    val baos = new ByteArrayOutputStream()
    val zos = new ZipOutputStream(new BufferedOutputStream(baos))

    try {
      fileList.map(file => {
        zos.putNextEntry(new ZipEntry(file.toPath.getFileName.toString))
        zos.write(FileUtils.readFileToByteArray(file))
        zos.closeEntry()
      })
    } finally {
      zos.close()
    }

    new ByteArrayInputStream(baos.toByteArray)
  }
}

Answer 1

您可以使用Java的管道机制，而不是使用ByteArrayOutputStream缓冲数组中的内容，然后将它们放入ByteArrayInputStream中。

这里有一个草图解决方案：

def zip() = Action {
  // Create Source that listens to an OutputStream
  // and pass it to `fileByteData` method.
  val zipSource: Source[ByteString, Unit] =
    StreamConverters
      .asOutputStream()
      .mapMaterializedValue(fileByteData)
  Ok.chunked(zipSource).withHeaders(
    CONTENT_TYPE -> "application/zip",
    CONTENT_DISPOSITION -> s"attachment; filename = test.zip")
}

// Send the file data, given an OutputStream to write to.
def fileByteData(os: OutputStream): Unit = {
  val fileList = List(
    new java.io.File("C:\\Users\\a.csv"),
    new java.io.File("C:\\Users\\b.csv")
  )

  val zos = new ZipOutputStream(os)
  val buffer: Array[Byte] = new Array[Byte](2048)
  try {
    for (file <- fileList) {
      zos.putNextEntry(new ZipEntry(file.toPath.getFileName.toString))
      val fis = new Files.newInputStream(file.toPath)
      try {
        @tailrec
        def zipFile(): Unit = {
          val bytesRead = fis.read(buffer)
          if (bytesRead == -1) () else {
            zos.write(buffer, 0, bytesRead)
            zipFile()
          }
        }
        zipFile()
      } finally fis.close()
      zos.closeEntry()
    }
  } finally {
    zos.close()
  }
}

这只是一个方法的概要。您还需要确保：-线程运行正常-filebytedata有望在发送线程的另一个线程上运行-错误处理正常-例如，如果服务器（如找不到文件）或客户端（提前断开连接）出现错误，所有流将正确关闭