java - 将生成的 Zip 的 Zip 提供给客户端

Question

我正在生成一个 ZIP 文件，其中包含一堆自动生成的 XML 文件。最近需求发生了变化，现在我必须生成几次 ZIP(带有 XML 数据的变体)并将它们直接提供给客户端，而不使用服务器中的实际文件。这就是我正在做的事情:

// [... servlet handling ... ]
    response.setContentType("application/zip");
    response.setHeader("Content-Disposition", "attachment;filename=cierresZ_a_tangonet" + java.time.LocalDate.now() + ".zip");
    // stream straight to client
    ServletOutputStream out = response.getOutputStream();
    ZipOutputStream zipped_out = new ZipOutputStream(out);
    for( each data block from db ){
    //CREATION AND PROCESSING OF XML FILES AS ZIP ENTRIES
       byte[] xmlBinData = xmlData.toString().getBytes();
       zipped_out.write(xmlBinData, 0, xmlBinData.length);
       zipped_out.flush();                                
     }
     zipped_out.finish();
     out.close();
}

尝试这样做但出现错误:

// [... servlet handling ... ]
  response.setContentType("application/zip");
  response.setHeader("Content-Disposition", "attachment;filename=cierresZ_a_tangonet" + java.time.LocalDate.now() + ".zip");
  // stream straight to client
  ServletOutputStream out = response.getOutputStream();
  for( each zip needed ){
    ZipOutputStream zipped_out = new ZipOutputStream(out);
    for( each data block from db ){
    //CREATION AND PROCESSING OF XML FILES AS ZIP ENTRIES
       byte[] xmlBinData = xmlData.toString().getBytes();
       zipped_out.write(xmlBinData, 0, xmlBinData.length);
       zipped_out.flush();                                
     }
     zipped_out.finish();
   }
   out.close();
}

编辑:对其进行了一些细微的更改，但给出了相同的错误

// [... servlet handling ... ]
  response.setContentType("application/zip");
  response.setHeader("Content-Disposition", "attachment;filename=cierresZ_a_tangonet" + java.time.LocalDate.now() + ".zip");
  // stream straight to client
  ServletOutputStream out = response.getOutputStream();
  ZipOutputStream zipped_outs = new ZipOutputStream(out);
  for( each zip needed ){
    //creates new zip inside big one
    ZipEntry zipFile = new ZipEntry(salaActual + ".zip");
    zipped_outs.putNextEntry(zipFile);

    //opens stream for this new zip
    ZipOutputStream zipped_out = new ZipOutputStream(zipped_outs);
    for( each data block from db ){
    //CREATION AND PROCESSING OF XML FILES AS ZIP ENTRIES
       byte[] xmlBinData = xmlData.toString().getBytes();
       zipped_out.write(xmlBinData, 0, xmlBinData.length);
       zipped_out.flush();                                
     } 
    //completes zip and closes it then goes for the next one
    byte[] zipBinData = zipFile.toString().getBytes();
    zipped_outs.write(zipBinData, 0, zipBinData.length);
    zipped_outs.flush();     
   }
   //closes the big zip filled with zips and returns
   zipped_outs.finish();
   out.close();
}

Answer 1

您需要finish()和flush()所有内部ZipOutputStream和外部ZipOutputStream。

基本上:

response.setContentType("application/zip");
response.setHeader("Content-Disposition", ...);
ZipOutputStream mainZip = new ZipOutputStream(response.getOutputStream());
for (each file to download) {
    ZipEntry zipEntry = new ZipEntry(fileName);
    mainZip.putNextEntry(zipEntry);
    if (file data available) {
        // write file data to mainZip here
    } else {
        ZipOutputStream subZip = new ZipOutputStream(mainZip);
        for (each subfile to download) {
            ZipEntry zipEntry = new ZipEntry(subFileName);
            subZip.putNextEntry(zipEntry);
            // write subfile data to subZip here
        }
        subZip.finish();
        subZip.flush(); // do not close
    }
}
mainZip.finish();
mainZip.close(); // flushes for you

请注意 mainZip 和 subZip 如何获取 zip 条目，并且都完成。您的代码中缺少这一点。