Question

我写了一个HTML代码来上传jsp.now中的文件我必须使用这个数据并使用jsp.can中的java程序来处理它你请给我一个如何实现这个的示例代码。

谢谢

Answer 1

我注意到你的代码。请按照以下步骤进行操作

下载apache commons-fileupload和apache commons-io的二进制发行版（FileUpload依赖于Commons IO，因此请确保在继续之前在类路径中的依赖项页面上提到了该版本。）
提取这两个发行版并在WEB-INF / lib文件夹下复制commons-fileupload-1.2.2.jar和commons-io-2.0.1.jar。
在Servlet中，您需要导入javax.servlet，javax.servlet.http，import java.io，java.util,org.apache.commons.fileupload，org.apache.commons.fileupload.servlet和org.apache.commons.fileupload.disk packages。

演示：

在WEB-INF / classes文件夹下创建一个servlet文件

package com.me;

import javax.servlet.ServletException;
import javax.servlet.http.*;
import java.io.IOException;
import java.io.File;
import java.util.List;
import java.util.Iterator;
import org.apache.commons.fileupload.FileItemFactory;
import org.apache.commons.fileupload.FileItem;
import org.apache.commons.fileupload.servlet.ServletFileUpload;
import org.apache.commons.fileupload.disk.DiskFileItemFactory;


public class FileUploadServlet extends HttpServlet
{
  public void doPost(HttpServletRequest request,
                     HttpServletResponse response) 
                         throws ServletException, IOException
   {
        boolean isMultipart = ServletFileUpload.isMultipartContent(request);
        java.io.PrintWriter pw=response.getWriter();
        final String SUCCESS="/success.jsp";
        final String FAIL="/fail.jsp";
        if(isMultipart) 
         {
          try
           {

             FileItemFactory factory = new DiskFileItemFactory();
             ServletFileUpload upload = new ServletFileUpload(factory);
             List items = upload.parseRequest(request);
             pw.print("total : " + items.size());
             Iterator iter = items.iterator();
             while (iter.hasNext()) 
             {
                FileItem item = (FileItem) iter.next();
             pw.print("total : " + item);
                if(!item.isFormField()) 
                  {
                     String fileName = item.getName();
                     String destFilePath=getServletContext().getRealPath("/upload/" + fileName);
                     File uploadedFile = new File(destFilePath);
                     item.write(uploadedFile);
                   }
              }
             request.setAttribute("message","file has been uploaded successfully!");
           }catch(Exception ex)
           {

              ex.printStackTrace();
              request.setAttribute("message","Cannot upload : " + ex.getMessage());
              getServletContext().getRequestDispatcher(FAIL).forward(request,response);
           }
           getServletContext().getRequestDispatcher(SUCCESS).forward(request,response);
        }
   }
}

在WEB-INF / web.xml中配置/注册servlet

<web-app>
 <servlet>
    <servlet-name>fileupload</servlet-name>
    <servlet-class>com.me.FileUploadServlet</servlet-class>
 </servlet>
 <servlet-mapping>
    <servlet-name>fileupload</servlet-name>
    <url-pattern>/fileupload</url-pattern>
 </servlet-mapping>
</web-app>

视图内容（upload.jsp）。

注意：使用“multipart / form-data”

<form method="post" enctype="multipart/form-data" action="fileupload">
  <input type="file" name="file"/>
  <input type="submit"/>
</form>

success.jsp和fail.jsp的内容

${message}

在upload的根目录下创建名为context的文件夹。

JSP版本（这不是推荐的方式）

upload1.jsp

<form method="post" enctype="multipart/form-data" 

action="showfile.jsp">
  <input type="file" name="file"/>
  <input type="submit"/>
</form>

showfile.jsp

<%@ page import="java.io.*" %>
<%@ page import="java.util.List" %>
<%@ page import="java.util.Iterator" %>
<%@ page import="org.apache.commons.fileupload.FileItemFactory" %>
<%@ page import="org.apache.commons.fileupload.FileItem" %>
<%@ page import="org.apache.commons.fileupload.servlet.ServletFileUpload" %>
<%@ page import="org.apache.commons.fileupload.disk.DiskFileItemFactory" %>


<%

 boolean isMultipart = ServletFileUpload.isMultipartContent(request);
 String content="";

 if(isMultipart) 
     {
        FileItemFactory factory = new DiskFileItemFactory();
        ServletFileUpload upload = new ServletFileUpload(factory);
        List items = upload.parseRequest(request);

        Iterator iter = items.iterator();
        while (iter.hasNext()) 
         {
            FileItem item = (FileItem) iter.next();
            if(!item.isFormField()) 
              {
                 BufferedInputStream buff=new BufferedInputStream(item.getInputStream());
                 byte []bytes=new byte[buff.available()];
                 buff.read(bytes,0,bytes.length);
                 content=new String(bytes);
              }
          }
     }
%>

Content of File:
<pre>
 <%=content%>
</pre>

如何在jsp中的java程序中获取和使用其上传文件的内容？

1 个答案: