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ZIP的数倍（XML数据有所不同）并将其直接提供给客户端，而无需使用服务器中的实际文件。我在做什么：

@Bean
public SaajSoapMessageFactory messageFactory() throws SOAPException {
        SaajSoapMessageFactory messageFactory = new SaajSoapMessageFactory();
        messageFactory.setMessageFactory(MessageFactory.newInstance(SOAPConstants.SOAP_1_2_PROTOCOL));
        return messageFactory;
    }

试图这样做，但给我错误：

// [... 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_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();
  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();
}

Answer 1

您需要finish()和flush()内部的所有ZipOutputStreams 和所有的外部ZipOutputStreams。

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都如何获取 zip条目，并且都完成了。这是您的代码中缺少的。

为客户提供生成的邮政编码的邮政编码

1 个答案: