如何打开WEB-INF中的文件? [英] How do I open a file located in WEB-INF?

问题描述

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.

My servlet lets user upload a file, I've created a button to view the uploaded file.

现在，单击该按钮时，我希望打开上载的文件.如何在JSP端或servlet.java端做到这一点?

now, on click of that button I want the uploaded file to open. How do I do this on the JSP side or servlet.java side?

它位于WEB-INF/Uploads/my.txt文件夹中.

it is located in WEB-INF/Uploads/my.txt folder.

=====================================编辑========= ===============================

=====================================EDIT=========================================

根据下面的答案，我已经修改了代码，并将其粘贴到此处以获取更多答案，

Based on answers below, I've modified my code and I'm pasting the same here for more answers,

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！

PS: This is not working, still looking for answers. Thank You!

推荐答案

尝试执行以下操作:

ServletContext context = getServletContext();
InputStream is = context.getResourceAsStream("/WEB-INF/Uploads/my.txt");

然后阅读如下所示的URL内容:

then read the URL content like the following :

  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.

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