Hi
On 23/06/13 20:50, Mandy Warren wrote:
Thanks very much for taking the time to reply..
Ok so I am running on a weblogic 10.3 container and it does seem that the
problem occurs when deploying to this environment but not when running on
Tomcat 7. On Tomcat I do get a 400 status code returned.
The 2 dodgy chars I am using in my url are \"
Even reading the link below I still don't really understand quite what might be
happening, would you mind explaining in a bit more detail so I can search for a
possible fix in the weblogic forums?
I guess you may want to ask how to prevent WebLogic decoding \" characters
Also, assuming weblogic has an issue, please could you explain how I write a
filter to trap the exception and convert the response code manually? I have
tried writing an Interceptor but the handleMessage method is never called
(perhaps I specified the phase incorrectly - I used PRE_LOGICAL?).
I was referring to a servlet filter. which is where you'd catch it
Cheers, Sergey
Many thanks
Mandy
On 23 Jun 2013, at 18:33, Sergey Beryozkin <[email protected]> wrote:
Hi
On 23/06/13 10:38, Andrei Shakirin wrote:
Hi,
Looks as defect for me. I will expect 400 Bad Request from CXF servlet in this
case.
Sergei, WDYT?
I think the reason for this particular exception is that the underlying
container decodes request URI unexpectedly, there was a Windows specific issue
to do with decoding %20, see
http://cxf.apache.org/docs/jaxrs-services-configuration.html#JAXRSServicesConfiguration-Tomcat,
but it could be caused by some other container specific issue.
One can always add a filter can catch exceptions from there, but that would
probably hide the container issue
Cheers, Sergey
Regards,
Andrei.
-----Original Message-----
From: Mandy Warren [mailto:[email protected]]
Sent: Samstag, 22. Juni 2013 17:01
To: [email protected]
Subject: How to intercept exceptions thrown by CXF servlet
I have developed a REST service using Apache CXF and notice that if I send
invalid characters in the URL, the CXF servlet throws back the following
exception before it gets to my code:
Servlet failed with Exception java.lang.IllegalArgumentException at
java.net.URI.create(URI.java:841) at
org.apache.cxf.transport.servlet.BaseUrlHelper.getBaseURL(BaseUrlHelper.j
ava:49) at
org.apache.cxf.transport.servlet.ServletController.getBaseURL(ServletContr
oller.java:73) at
org.apache.cxf.transport.servlet.ServletController.updateDestination(Servle
tController.java:82) at
org.apache.cxf.transport.servlet.ServletController.invoke(ServletController.j
ava:162) at
org.apache.cxf.transport.servlet.ServletController.invoke(ServletController.j
ava:137) at
org.apache.cxf.transport.servlet.CXFNonSpringServlet.invoke(CXFNonSpring
Servlet.java:158) at
org.apache.cxf.transport.servlet.AbstractHTTPServlet.handleRequest(Abstra
ctHTTPServlet.java:239) at
org.apache.cxf.transport.servlet.AbstractHTTPServlet.doGet(AbstractHTTPS
ervlet.java:164) at
javax.servlet.http.HttpServlet.service(HttpServlet.java:707) at
org.apache.cxf.transport.servlet.AbstractHTTPServlet.service(AbstractHTTPS
ervlet.java:215)
This results in a 500 status code being returned to the client.
I would ideally like to intercept this exception and return a 400 Bad Request
status code to the client but I am unable to work out how to do this.
Any help much appreciated!
Many thanks
--
Sergey Beryozkin
Talend Community Coders
http://coders.talend.com/
Blog: http://sberyozkin.blogspot.com
