How to get the offending uri when handling a 404 error?

[Joseph Shraibman]

I put this in my web.xml:
<error-page>
<error-code>404</error-code>
<location>/404.jsp</location>
</error-page>
... and this in 404.jsp:
ErrorData ed = pageContext.getErrorData();
if (ed != null) url = ed.getRequestURI() ;
but it throws:
StandardWrapperValve[jsp]: Servlet.service() for servlet jsp threw exception
java.lang.NullPointerException
java.lang.NullPointerException
        at javax.servlet.jsp.PageContext.getErrorData(PageContext.java:514)
        at org.apache.jsp._404_jsp._jspService(_404_jsp.java:102)
        at 
org.apache.jasper.runtime.HttpJspBase.service(HttpJspBase.java:94)
        at javax.servlet.http.HttpServlet.service(HttpServlet.java:810)
        at 
org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:298)
        at 
org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:292)
        at 
org.apache.jasper.servlet.JspServlet.service(JspServlet.java:236)

(this is with tomcat 5.0.24)
So how can I get the url the the user tried to request?
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