我的servlet允许用户上传文件,我创建了一个按钮来查看上传的文件。
现在,单击该按钮,我希望打开上载的文件。如何在JSP端或servlet.java端执行此操作?
它位于WEB-INF / Uploads / my.txt文件夹中。
根据下面的答案,我已经修改了代码,并在此处粘贴了相同的代码以获取更多答案,
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException { ServletContext context = getServletContext(); String path = context.getRealPath("/u/poolla/workspace/FirstServlet/WebContent/WEB-INF/Uploads/Config.txt"); FileReader reader = new FileReader(path); BufferedReader br = new BufferedReader(reader); String firstline = br.readLine(); System.out.println(firstline); }
PS:这不起作用,仍在寻找答案。谢谢!
尝试执行以下操作:
ServletContext context = getServletContext(); InputStream is = context.getResourceAsStream("/WEB-INF/Uploads/my.txt");
然后读取网址内容,如下所示:
BufferedReader br = new BufferedReader(new InputStreamReader( is)); int value=0; // reads to the end of the stream while((value = br.read()) != -1) { // converts int to character char c = (char)value; // prints character System.out.println(c); }
请给我一些反馈
希望有帮助。