如何将JSP正确放入WEB-INF中？
我正在使用 MVC，并且希望将 JSP 页面放在 WEB-INF 中以避免直接访问它。我在 Web 内容的 jsp 文件夹中有一个 index.jsp 页面和其他页面，并且它可以工作。它看起来像这样：
-Web Content
-index.jsp
-jsp
--main_read.jsp
--...
顺便说一句，index.jsp 是我的登录页面，无论用户是否登录，在我使用的控制器中
RequestDispatcher dispatcher =
getServletContext().getRequestDispatcher("jsp/main_read.jsp");
dispatcher.forward(request, response);
我工作得很好，但是当我尝试将 JSP 放入 WEB-INF 时，它失败了：
-Web Content
-index.jsp
-WEB-INF
--jsp
---jsp
----main_read.jsp
----...
并给出了像这样的错误
HTTP Status 404 - /Libruary/jsp/main_read.jsp
type Status report
message /Libruary/jsp/main_read.jsp
description The requested resource (/Libruary/jsp/main_read.jsp) is not available.
Apache Tomcat/6.0.26
可能问题出在页面路径中，我在 dispatcher.forward 中编写，但无论如何，请帮助我。
I'm using MVC and want to put my JSP pages in WEB-INF to avoid direct access to it. I have an index.jsp page and other pages in jsp folder in Web Content and it works. It looks like this:
-Web Content
-index.jsp
-jsp
--main_read.jsp
--...
By the way, index.jsp is my login page and whether user is logged, in controller I use
RequestDispatcher dispatcher =
getServletContext().getRequestDispatcher("jsp/main_read.jsp");
dispatcher.forward(request, response);
I works perfect, but when I'm trying to put my JSP in WEB-INF it fails:
-Web Content
-index.jsp
-WEB-INF
--jsp
---jsp
----main_read.jsp
----...
And gives an error like this
HTTP Status 404 - /Libruary/jsp/main_read.jsp
type Status report
message /Libruary/jsp/main_read.jsp
description The requested resource (/Libruary/jsp/main_read.jsp) is not available.
Apache Tomcat/6.0.26
Probably the problem is in the page path, I write in dispatcher.forward, but anyway, help me please.
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您似乎传递了路径
/jsp/main_read.jsp，并且JSP位于/WEB-INF/jsp/jsp/main_read.jsp中。显然，路径不匹配。将正确的路径传递给getRequestDispatcher()：/WEB-INF/jsp/jsp/main_read.jsp。javadoc 说：
You seem to pass the path
/jsp/main_read.jsp, and the JSP is in/WEB-INF/jsp/jsp/main_read.jsp. Obviously, the paths don't match. Pass the correct path togetRequestDispatcher():/WEB-INF/jsp/jsp/main_read.jsp.The javadoc says: