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From "Shapira, Yoav" <Yoav.Shap...@mpi.com>
Subject RE: How to get the offending uri when handling a 404 error?
Date Wed, 02 Jun 2004 12:50:26 GMT

Hi,
Does your 404.jsp have the isErrorPage directive set to true?

Yoav Shapira
Millennium Research Informatics


>-----Original Message-----
>From: Joseph Shraibman [mailto:jks@selectacast.net]
>Sent: Tuesday, June 01, 2004 7:57 PM
>To: tomcat-user@jakarta.apache.org
>Subject: How to get the offending uri when handling a 404 error?
>
>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.j
ava:
>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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