How do I create zip file in Servlet for download?
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2022-04-24 14:19:32
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原文链接:https://kodejava.org/how-do-i-create-zip-file-in-servlet-for-download/
The example below is a servlet that shows you how to create a zip file and send the generated zip file for user to download. The compressing process is done by the zipFiles method of this class.
For a servlet to work you need to configure it in the web.xml file of your web application which can be found after the code snippet below.
package org.kodejava.example.servlet;
import javax.servlet.ServletException;
import javax.servlet.ServletOutputStream;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import java.io.*;
import java.util.zip.ZipEntry;
import java.util.zip.ZipOutputStream;
public class ZipDownloadServlet extends HttpServlet {
public static final String FILE_SEPARATOR = System.getProperty("file.separator");
protected void doPost(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
doGet(request, response);
}
protected void doGet(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
try {
//
// The path below is the root directory of data to be
// compressed.
//
String path = getServletContext().getRealPath("data");
File directory = new File(path);
String[] files = directory.list();
//
// Checks to see if the directory contains some files.
//
if (files != null && files.length > 0) {
//
// Call the zipFiles method for creating a zip stream.
//
byte[] zip = zipFiles(directory, files);
//
// Sends the response back to the user / browser. The
// content for zip file type is "application/zip". We
// also set the content disposition as attachment for
// the browser to show a dialog that will let user
// choose what action will he do to the sent content.
//
ServletOutputStream sos = response.getOutputStream();
response.setContentType("application/zip");
response.setHeader("Content-Disposition", "attachment; filename="DATA.ZIP"");
sos.write(zip);
sos.flush();
}
}
catch (Exception e) {
e.printStackTrace();
}
}
/**
* Compress the given directory with all its files.
*/
private byte[] zipFiles(File directory, String[] files) throws IOException {
ByteArrayOutputStream baos = new ByteArrayOutputStream();
ZipOutputStream zos = new ZipOutputStream(baos);
byte bytes[] = new byte[2048];
for (String fileName : files) {
FileInputStream fis = new FileInputStream(directory.getPath() +
ZipDownloadServlet.FILE_SEPARATOR + fileName);
BufferedInputStream bis = new BufferedInputStream(fis);
zos.putNextEntry(new ZipEntry(fileName));
int bytesRead;
while ((bytesRead = bis.read(bytes)) != -1) {
zos.write(bytes, 0, bytesRead);
}
zos.closeEntry();
bis.close();
fis.close();
}
zos.flush();
baos.flush();
zos.close();
baos.close();
return baos.toByteArray();
}
}
The web.xml configuration:
<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns="http://java.sun.com/xml/ns/javaee"
xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
id="WebApp_ID" version="2.5">
<servlet>
<servlet-name>ZipDownloadServlet</servlet-name>
<servlet-class>org.kodejava.example.servlet.ZipDownloadServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>ZipDownloadServlet</servlet-name>
<url-pattern>/zipservlet</url-pattern>
</servlet-mapping>
</web-app>