CODEWITHSUNDEEP
javajspservlets
Jeremy Logan
Jeremy Logan

Reputation: 47514

How do you detect the URL in a Java Servlet when forwarding to JSP?

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());
    }
}

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>

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);
    }
}

then when I check the value of getPathInfo() it now reports "whatever.jsp" instead of "foo".

  1. Why has this changed before it's been forwarded to the JSP?
  2. How can I detect what URL the user's looking for?

EDIT: Just in case it matters this is on Google App Engine. Don't think it should though.

Upvotes: 4

Views: 11210

Answers (2)

BalusC
BalusC

Reputation: 1109532

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").

Upvotes: 16

Tom
Tom

Reputation: 44881

The URL the user (browser) requested can be accessed from the request by:

request.getRequestURL()

Alternatively the request has a whole bunch of accessors to get the various pieces of the URL as well as those on ServletRequest.

To redirect to a different URL change the response rather than the request:

response.sendRedirect(theURLToRedirectTo)

Upvotes: 2

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