问题描述
我有一个看起来像这样的servlet:
I have a servlet that looks something like this:
public class ExampleServlet extends HttpServlet {
public void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
response.getWriter().println(request.getPathInfo());
}
}
带有web.xml映射,如:
with a web.xml mapping like:
<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......如果我去它打印/ foo。但是,如果我将servlet更改为转发到JSP文件:
and it gives me exactly what I expect... If I go to http://localhost:8080/example/foo it prints "/foo". However, if I change the servlet to forward to a JSP file:
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()的值时现在报告whatever.jsp而不是foo。
then when I check the value of getPathInfo() it now reports "whatever.jsp" instead of "foo".
- 为什么在转发到JSP之前这已经改变了?
- 如何检测用户正在寻找的网址?
编辑:万一重要,这是在Google App Engine上。不要认为应该这样。
Just in case it matters this is on Google App Engine. Don't think it should though.
推荐答案
这个问题模糊不清(servlet是否会再次呼唤每个前锋? ),但听起来很像你需要 request.getAttribute(javax.servlet.forward.request_uri)
。
The question is vague and ambiguous (is the servlet calling itself on every forward again?), but it much sounds like that you need request.getAttribute("javax.servlet.forward.request_uri")
.
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