通过HttpResponse Java

Question

所以我从一个数据库（各种mimetypes）中抓取一组blob，然后尝试将它们压缩起来，供用户通过http响应下载。我可以进行下载，但是当我尝试打开下载的zip文件时，它说“存档格式未知或已损坏”。我已经尝试了以下代码与application / zip，application / octet-stream和application / x-zip-compressed，但我开始假设问题在于我如何添加文件。我也使用Java 7和Grails 2.2.4。

对此的任何帮助将不胜感激。谢谢！

  final ZipOutputStream out = new ZipOutputStream(new FileOutputStream("test.zip"));


        for (Long id : ids){

            Object[] stream = inlineSamplesDataProvider.getAttachmentStream(id);


            if (stream) {

                String fileName = stream[0]
                String mimeType = (String) stream[1];
                InputStream inputStream = stream[2]
                byte[] byteStream = inputStream.getBytes();

                ZipEntry zipEntry = new ZipEntry(fileName)
                out.putNextEntry(zipEntry);
                out.write(byteStream, 0, byteStream.length);
                out.closeEntry();
            }
        }

        out.close();
        response.setHeader("Content-Disposition", "attachment; filename=\"" + "test.zip" + "\"");
        response.setHeader("Content-Type", "application/zip");
        response.outputStream << out;
        response.outputstream.flush();

Answer 1

我在这里找到答案：Returning ZipOutputStream to browser

好吧，最终为我工作的是将ZipOutputStream转换为ByteArrayOutputStream并将其作为byte []写入响应：

/blog/2015/10/20/sample-post/

感谢所有帮助过的人！