How to get the original request url from a servlet/jsp after multiple servlet forwards

Question

I am working on a cruise booking app using struts/tiles that uses multiple internal servlet/jsp forwards to reach the right jsp for display. But, once you reach the final jsp that is used to render the page, the ${pageContext.request.requestURL} call in that jsp returns the path of this jsp.

For example

1. Original request: /booking/getCruiseDetails
2. gets forwarded to: /booking/validateCruiseDeteails.jsp
3. gets forwarded to: /booking/validateUser.jsp
4. finally gets forwarded to: /booking/showCruiseDetails.jsp

So, in /booking/showCruiseDetails.jsp when I call ${pageContext.request.requestURL} I get /booking/showCruiseDetails.jsp

How do you get the the original (client made) request url from a jsp that has been reached through multiple forwards. I did find the following posts on stackoverflow that hint at the solution here and here, but they don't address how you would go about finding the original request url after multiple forwards have occurred.

Answer 1

I found a better answer in this post [ How do you detect the URL in a Java Servlet when forwarding to JSP? ]

On the target JSP use:

request.getAttribute("javax.servlet.forward.request_uri")

To find out what the original URL was.

It doesn't require you to take any extra steps on the forwarding servlet

Answer 2

You can use a filter to putting origin address to request attribute and then read it from jsp

Filter mapped to /booking/* execute:

request.setAttribute("origin", request.getRequestURL());

Jsp:

${pageContext.request.attribute["origin"]}

This works because filter has set REQUEST dispatcher by default. It means filter executes only for direct client requests not for forwarding/including

Answer 3

${requestScope["javax.servlet.forward.request_uri"]}

or with single quotes

${requestScope['javax.servlet.forward.request_uri']}

Answer 4

Consider using servlet filters instead to validate information. This means you can avoid your validation forwarding and just stay in a single JSP file.

Answer 5

You can show it without using a bean reference with the following:

<h:outputText value="#{requestScope['javax.servlet.forward.request_uri']}" />

Of course, you need to map the 404 page in your web.xml file though.

<error-page>
    <error-code>404</error-code>
    <location>/xhtml/pg/error/404.xhtml</location>
</error-page>

Answer 6

Same as @Lenny Markus but using the provided constant in the Request Dispatcher class.

request.getAttribute(RequestDispatcher.FORWARD_REQUEST_URI)