java实现文件上传(使用FromData)
一、使用FromData将表单对象序列化后,使用request.getInputStream()来获取数据
1、表单代码如下
<form id="user-info" method="post" action="upload" > <input type="file" name="file" multiple> <input type="text" name="username"> <input type="text" name="password"> </form>
<button id="sub">提交</button>
2、js的代码
<script> function createXHR(){ return new XMLHttpRequest(); } var sub = document.getElementById("sub"); sub.onclick=function(){ var xhr = createXHR(); var form = document.getElementById("user-info");//获取上边的表单 xhr.open("post","upload",true); xhr.send(new FormData(form)); } </script>
3、由于是序列化的表单所以java后端，不能使用request.getParameter()获取数据，需要使用request.getInputStream();获取数据
首先我们看到获取的数据
InputStream in = request.getInputStream(); BufferedReader br = new BufferedReader(new InputStreamReader(in)); String str = ""; while((str=br.readLine())!=null){ System.out.println(str); }
可以看到读取到这些数据
4、接着使用apache的上传文件框架进行上传
实现效果
以下是代码
package com.wangyang.servlet; import java.io.File; import java.io.IOException; import java.util.List; import javax.servlet.ServletException; import javax.servlet.annotation.WebServlet; import javax.servlet.http.HttpServlet; import javax.servlet.http.HttpServletRequest; import javax.servlet.http.HttpServletResponse; import org.apache.commons.fileupload.FileItem; import org.apache.commons.fileupload.FileUploadException; import org.apache.commons.fileupload.disk.DiskFileItemFactory; import org.apache.commons.fileupload.servlet.ServletFileUpload; @WebServlet("/upload") public class Upload extends HttpServlet { private static final long serialVersionUID = 1L; protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException { request.setCharacterEncoding("utf-8"); response.setContentType("text/html;charset=utf-8"); /* InputStream in = request.getInputStream(); BufferedReader br = new BufferedReader(new InputStreamReader(in)); String str = ""; while((str=br.readLine())!=null){ System.out.println(str); } */ System.out.println("[-------------------------------------------------------]"); String filepath=request.getServletContext().getRealPath("/")+"upload/"; File file = new File(filepath); if(!file.exists()) { file.mkdir(); } DiskFileItemFactory factory = new DiskFileItemFactory(); ServletFileUpload upload = new ServletFileUpload(factory); try { List<FileItem> items= upload.parseRequest(request); for(FileItem item: items) { System.out.println(filepath+item.getName()); if(!item.isFormField()) { item.write(new File(filepath+item.getName())); } if(item.isFormField()){ System.out.println(item.getString()); System.out.println(item.getFieldName()); } } } catch (FileUploadException e) { e.printStackTrace(); } catch (Exception e) { // TODO Auto-generated catch block e.printStackTrace(); } response.sendRedirect("index.jsp"); } }
<%@ page language="java" contentType="text/html; charset=UTF-8" pageEncoding="UTF-8"%> <!DOCTYPE html> <html> <head> <meta charset="UTF-8"> <meta name="viewport" content="width=device-width, initial-scale=1.0"> <title>Insert title here</title> </head> <body> <form id="user-info" method="post" action="upload" > <input type="file" name="file" multiple> <input type="text" name="username"> <input type="text" name="password"> </form> <button id="sub">提交</button> <script> function createXHR(){ return new XMLHttpRequest(); } var sub = document.getElementById("sub"); sub.onclick=function(){ var xhr = createXHR(); var form = document.getElementById("user-info"); xhr.open("post","upload",true); xhr.send(new FormData(form)); } </script> </body> </html>
这里没有进行细节上的编程，只是为了说明简单原理
