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关于file控件上传的文件路径有关问题

热度:76   发布时间:2016-04-22 01:20:56.0
关于file控件上传的文件路径问题
file控件点击浏览按钮,获取他选择文件的路径,然后根据这个路径读取文件里面的内容,将他显示到文本框里,怎么做啊求教

实现的功能是 点击浏览的时候你选择一个文件,这时候这个file控件里就有了选择的文件的路径,想通过这个路径,把选择的文件的内容读取到 文本框中,这个要怎么实现,求解啊~~

------解决方案--------------------
Java code
  public class ImageServlet extends HttpServlet {            /**     * The doGet method of the servlet. <br>     *     * This method is called when a form has its tag value method equals to get.     *      * @param request the request send by the client to the server     * @param response the response send by the server to the client     * @throws ServletException if an error occurred     * @throws IOException if an error occurred     */    public void doGet(HttpServletRequest request, HttpServletResponse response)            throws ServletException, IOException {        this.doPost(request, response);    }    /**     * The doPost method of the servlet. <br>     *     * This method is called when a form has its tag value method equals to post.     *      * @param request the request send by the client to the server     * @param response the response send by the server to the client     * @throws ServletException if an error occurred     * @throws IOException if an error occurred     */    public void doPost(HttpServletRequest request, HttpServletResponse response)            throws ServletException, IOException {        response.setContentType("text/html");        response.setCharacterEncoding("utf-8");        MultipartRequest mr = null;        int maxPostSize = 1*100*1024*1024;        String dir = getServletConfig().getServletContext().getRealPath("/uploadImage");        mr = new MultipartRequest(request,dir,maxPostSize,"utf-8");        Enumeration files = mr.getFileNames();        String fileName ="";        String path = "";        while(files.hasMoreElements()){                        fileName = files.nextElement().toString();            path = mr.getFilesystemName(fileName);            File f = mr.getFile(fileName);                        if(f==null){                throw new ServletException("file is null");            }            Date dt = new Date(System                      .currentTimeMillis());              SimpleDateFormat fmt = new SimpleDateFormat(              "yyyyMMddHHmmssSSS");              String time = fmt.format(dt);             String newName = time+f.getName().substring(path.lastIndexOf('.'));            File newFile = new File(dir+"/"+newName);            f.renameTo(newFile);            System.out.println("newName:"+newName);            System.out.println("filename:"+newFile.getName());            System.out.println("path:"+newFile.getPath());        }        PrintWriter out = response.getWriter();        out.print("上传成功!");    }}
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