Question

我需要从servlet动态生成包含数据的ZIP文件。我将ServletOutputSream传递给ZipOutputStream构造函数，然后它正确地将ZIP发送到客户端。当ZIP结果很大时，servlet会回答很多。我想首先创建ZIP并在发送之后。

我创建了一个包装ZipOutputStream的帮助器类。

我需要在调用flush方法时，将立即发送已生成的ZIP片段。我怎么能这样做？

public class ZIPResponse {

    private ZipOutputStream out;
    private HttpServletResponse response;
    private boolean isFirstEntry;
    private ConnectionManager con;

    public ZIPResponse(ConnectionManager con, OutputStream out) throws IOException {
        this.out = new ZipOutputStream(out);
        isFirstEntry = true;
        this.con = con;
    }

    public ZIPResponse(ConnectionManager con, HttpServletResponse response) throws IOException {
        this(con, response.getOutputStream());
        this.response = response;
        response.setCharacterEncoding("ISO-8859-1");
        sendNewEntry("con");
        flush();
    }

    public void sendNewEntry(String dataName) throws IOException {
        if (isFirstEntry) {
            isFirstEntry = false;
        } else {
            out.closeEntry();
        }
        out.putNextEntry(new ZipEntry(dataName));
    }

    public void sendData(String dataName, String data) throws IOException {
        sendNewEntry(dataName);
        sendData(data);
    }

    public void sendData(byte[] bytes) throws IOException {
        out.write(bytes);
        flush();
    }

    public void sendData(String data) throws IOException {
        sendData(data.getBytes());
    }

    public void sendAndCloseData(String dataName, String data) throws IOException {
        sendNewEntry(dataName);
        sendData(data);
        close();
    }

    private void sendDataFile(String dataName, DataTransformer dt)
            throws IOException, SQLException {
        sendNewEntry(dataName);
        dt.sendResultSet(this);
    }

    public void sendDataFile(String dataName, ResultSet data, String[] columnsRestriction) throws IOException,
            SQLException {
        RestrictedDataTransformer dt = new RestrictedDataTransformer(data, columnsRestriction);
        sendDataFile(dataName, dt);
    }

    public void sendDataFile(String dataName, ResultSet data) throws IOException,
            SQLException {
        DataTransformer dt = new DataTransformer(data);
        sendDataFile(dataName, dt);
    }

    public void sendDataFile(String dataName, PreparedStatement p) throws
            SQLException, IOException {
        ResultSet data = p.executeQuery();
        sendDataFile(dataName, data);
    }


    public void sendDataFile(String dataName, String SQL, String[] columnsRestricted)
            throws DBException, SQLException, IOException {
        ResultSetStatement st = new ResultSetStatement(con, SQL);
        ResultSet valores = st.executeQuery();
        try {
            sendDataFile(dataName, valores, columnsRestricted);
        } finally {
            valores.close();
        }
    }

    public void sendDataFile(String dataName, String SQL) throws DBException, SQLException, IOException {
        ResultSetStatement st = new ResultSetStatement(con, SQL);
        ResultSet valores = st.executeQuery();
        try {
            sendDataFile(dataName, valores);
        } finally {
            valores.close();
        }
    }

    public void sendAndCloseDataFile(String dataName, ResultSet data) throws
            SQLException, IOException {
        sendDataFile(dataName, data);
        close();
    }

    public void sendAndCloseDataFile(String dataName, String SQL) throws DBException, SQLException, IOException {
        sendDataFile(dataName, SQL);
        close();
    }

    public void close() throws IOException {
        sendNewEntry("OK");
        out.closeEntry();
        out.close();
        flush();
    }

    public void flush() throws IOException {       
      out.flush();
      response.flushBuffer();
    }
}

Answer 1

如果您的客户对收到的内容非常灵活，GZIP将更适合流媒体压缩。

Answer 2

这是我的实施：

String realpath = getServletContext().getRealPath("/");
    response.setContentType("application/zip");
    response.setHeader("Content-Disposition","attachment; filename="+fi.replace('/', '-')+"_"+ff.replace('/', '-')+".zip");
ServletOutputStream out = null;
ZipOutputStream zipfile = null;
try{
                List<Object[]> cfdis = /*my hibernate criteria source, your Database?*/

        out = response.getOutputStream();

        zipfile = new ZipOutputStream(out);
        ZipEntry zipentry = null;

        for(Object[] cfdi:cfdis){

            zipentry = new ZipEntry(cfdi[1].toString()+".xml");
            zipfile.putNextEntry(zipentry);

            InputStream in = new FileInputStream(new File(realpath+cfdi[0].toString()));


            byte[] bytes = new byte[FILEBUFFERSIZE];
            int bytesRead;
            while ((bytesRead = in.read(bytes)) != -1) {
                zipfile.write(bytes, 0, bytesRead);
            }
        }
}catch(...){}
finally{
   zipfile.close();
   out.close();
}

我希望你觉得这很有用。

Answer 3

我想你的问题是在完成zip创建之前将zip内容发送到响应。是的，你可以通过缓冲来做到这一点。

对于缓冲，请按照Streaming large files in a java servlet

进行操作

立即邮寄ZIP

3 个答案: