To get the 404 url use the getRequestURL method of HttpServletRequest:
request.getRequestURL()
That returns a string representation of the URL that generated the request.
Your might also consider setting up a filter that recognizes and logs those bad requests.
Hope that helps,
Mr. Ariel S. Valentin mailto: [EMAIL PROTECTED]
From: "Mike Curwen" <[EMAIL PROTECTED]> Reply-To: "Tomcat Users List" <[EMAIL PROTECTED]> To: "'Tomcat Users List'" <[EMAIL PROTECTED]> Subject: RE: How to get the offending uri when handling a 404 error? Date: Wed, 2 Jun 2004 09:37:48 -0500
I got a related question:
Even if you fail to set the isErrorPage directive, would you still be able to access the information through the request attributes?
such as: javax.servlet.error.request_uri
> -----Original Message----- > From: Shapira, Yoav [mailto:[EMAIL PROTECTED] > Sent: Wednesday, June 02, 2004 7:50 AM > To: Tomcat Users List > Subject: RE: How to get the offending uri when handling a 404 error? > > > > Hi, > Does your 404.jsp have the isErrorPage directive set to true? > > Yoav Shapira > Millennium Research Informatics > > > >-----Original Message----- > >From: Joseph Shraibman [mailto:[EMAIL PROTECTED] > >Sent: Tuesday, June 01, 2004 7:57 PM > >To: [EMAIL PROTECTED] > >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(JspServle > tWrapper.j > ava: > >298) > > at > >org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServle > t.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? > > > >--------------------------------------------------------------------- > >To unsubscribe, e-mail: [EMAIL PROTECTED] > >For additional commands, e-mail: [EMAIL PROTECTED] > > > > > This e-mail, including any attachments, is a confidential > business communication, and may contain information that is > confidential, proprietary and/or privileged. This e-mail is > intended only for the individual(s) to whom it is addressed, > and may not be saved, copied, printed, disclosed or used by > anyone else. If you are not the(an) intended recipient, > please immediately delete this e-mail from your computer > system and notify the sender. Thank you. > > > --------------------------------------------------------------------- > To unsubscribe, e-mail: [EMAIL PROTECTED] > For additional commands, e-mail: [EMAIL PROTECTED] >
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