在转发到JSP时,如何检测Java Servlet中的URL？

Question

我有一个看起来像这样的servlet:

public class ExampleServlet extends HttpServlet {
    public void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        response.getWriter().println(request.getPathInfo());
    }
}

使用web.xml映射,如:

<servlet>
    <servlet-name>example</servlet-name>
    <servlet-class>com.example.ExampleServlet</servlet-class>
</servlet>
<servlet-mapping>
    <servlet-name>example</servlet-name>
    <url-pattern>/example/*</url-pattern>
</servlet-mapping>

它给了我我期望的结果......如果我去http:// localhost:8080/example/foo它打印"/ foo".但是,如果我将servlet更改为转发到JSP文件:

public class ExampleServlet extends HttpServlet {
    public void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        // do something here to check the value of request.getPathInfo()
        request.getRequestDispatcher("whatever.jsp").forward(request, response);
    }
}

然后当我检查getPathInfo()的值时,它现在报告"whatever.jsp"而不是"foo".

1. 为什么在将其转发到JSP之前已经改变了？
2. 如何检测用户正在寻找的URL？

编辑:以防万一,这是在谷歌应用引擎上.不要以为应该这样.

Answer 1

问题是含糊不清的(servlet是否会在每个前锋上再次调用？),但听起来很像你需要的request.getAttribute("javax.servlet.forward.request_uri").