I checked out the spec section 8.4.2 Forwarded Request Parameters.
It does seem to me that it implies that the parameters is where one
would get the original URI from.
However, there are still 2 problems that I can see:
1) The API says: "URL the client used to make the request"
2) 8.4.2 says that these parameters are NOT to be set if you obtain
the RequestDispatcher object using getNamedDispatcher() method. How
would you get at the original URI/URL then (without custom coding)?
Yaakov.
On 4/15/05, Trond G. Ziarkowski <[EMAIL PROTECTED]> wrote:
> Hi,
>
> "pretty sure" isn't always good enough ;) To get the uri that forwarded
> to the jsp you need to use
> <%=request.getAttribute("javax.servlet.forward.request_uri")%>. Check
> out the servlet 2.4 spec section 8.4 for more info.
>
>
> Trond
>
> Yaakov Chaikin wrote:
>
> >Hi,
> >
> >Using Tomcat 5.5.7 (tried Tomcat 5.5.9 with the same results) on Windows XP.
> >
> >When forwarding to a JSP page that is located in
> >/WEB-INF/jsp/success.jsp and calling:
> ><%= request.getRequestURI() %> inside the success.jsp page, the result I get
> >is:
> >/WEB-INF/jsp/success.jsp
> >
> >I am pretty sure that according to the API, this is the wrong result.
> >It should have returned the URI of the **request**, not the path to
> >the resource.
> >
> >Is this a known bug or there is some weird Tomcat setting that I need
> >to change. I ran this on Tomcat without changing any of the original
> >settings.
> >
> >BTW, the same is true of request.getRequestURL(). It returns (peculiar
> >enough):
> >http://localhost:8080/WEB-INF/jsp/success.jsp
> >
> >Thanks,
> >Yaakov.
> >
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> >
> >
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