在播放框架中下载动态创建的 zip 文件

Question

您好，我正在尝试编写一个可以下载多个文件的播放框架服务。我即时创建了多个文件的 zip，但我不确定如何在 Play Framework 中将其作为响应发送，我将展示我到目前为止所做的工作。

 public Result download() {

     String[] items = request().queryString().get("items[]");
        String toFilename = request().getQueryString("toFilename");

        ByteArrayOutputStream baos = new ByteArrayOutputStream();
        try (ZipOutputStream zos = new ZipOutputStream(new BufferedOutputStream(baos))) {
            for (String item : items) {
                Path path = Paths.get(REPOSITORY_BASE_PATH, item);
                if (Files.exists(path)) {
                    ZipEntry zipEntry = new ZipEntry(path.getFileName().toString());
                    zos.putNextEntry(zipEntry);
                    byte buffer[] = new byte[2048];
                    try (BufferedInputStream bis = new BufferedInputStream(Files.newInputStream(path))) {
                        int bytesRead = 0;
                        while ((bytesRead = bis.read(buffer)) != -1) {
                            zos.write(buffer, 0, bytesRead);
                        }
                    } finally {
                        zos.closeEntry();
                    }
                }
            }

            response().setHeader("Content-Type", "application/zip");
            response().setHeader("Content-Disposition", "inline; filename=\"" + MimeUtility.encodeWord(toFilename) + "\"");


//I am confused here how to output the response of zip file i have created
//I tried with the `baos` and with `zos` streams but not working

            return ok(baos.toByteArray());

        } catch (IOException e) {
            LOG.error("copy:" + e.getMessage(), e);
            return ok(error(e.getMessage()).toJSONString());
        }

        return null;
    }

我尝试使用return ok(baos.toByteArray()); 发送响应我能够下载文件，但是当我打开下载的文件时它给我错误An error occurred while loading the archive。

Answer 1

感谢您的帮助！我做了两个额外的更改，因此它适用于 scala playframework 2.5.x

1. 而不是 return ok(baos.toByteArray()) ， 使用Ok.chunked(StreamConverters.fromInputStream(fileByteData))

2. FileUtils.readFileToByteArray(file) 不是逐字节读取文件，而是在这里非常有帮助。

附件是我的代码的完整版本。

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 2

如果有人想知道最终代码是什么，我将添加此答案：

        String[] items = request().queryString().get("items[]");
        String toFilename = request().getQueryString("toFilename");

        ByteArrayOutputStream baos = new ByteArrayOutputStream();
        try (ZipOutputStream zos = new ZipOutputStream(new BufferedOutputStream(baos))) {
            for (String item : items) {
                Path path = Paths.get(REPOSITORY_BASE_PATH, item);
                if (Files.exists(path)) {
                    ZipEntry zipEntry = new ZipEntry(path.getFileName().toString());
                    zos.putNextEntry(zipEntry);
                    byte buffer[] = new byte[2048];
                    try (BufferedInputStream bis = new BufferedInputStream(Files.newInputStream(path))) {
                        int bytesRead = 0;
                        while ((bytesRead = bis.read(buffer)) != -1) {
                            zos.write(buffer, 0, bytesRead);
                        }

                    } finally {
                        zos.closeEntry();
                    }
                }
            }

            zos.close(); //closing the Zip

            response().setHeader("Content-Type", "application/zip");
            response().setHeader("Content-Disposition", "attachment; filename=\"" + MimeUtility.encodeWord(toFilename) + "\"");
            return ok(baos.toByteArray());

        } catch (IOException e) {
            LOG.error("copy:" + e.getMessage(), e);
            return ok(error(e.getMessage()).toJSONString());
        }

Answer 3

您需要关闭 zip 文件。添加所有条目后，执行：zos.close()

附带说明，我建议将 zip 文件写入磁盘，而不是将其保存在内存缓冲区中。然后您可以使用return ok(File content, String filename) 将其内容发送给客户端。