Fabbris Pierluigi wrote:
Hello,
I've this Exception in the Log:
7-ago-2007 10.38.05 org.apache.catalina.core.StandardWrapperValve invoke
GRAVE: Servlet.service() for servlet invoker threw exception
java.util.MissingResourceException: Can't find bundle for base name
LocalStrings, locale it_IT
Hi Gregor,
I am not sure about the definite solution, but as you
got no other answers, try this:
Drop an additional log4j.jar in the webapps lib folder.
Only this, IMHO, makes the webapp load another instance
of log4j so it can be configured independently.
The log4j.xml would then be searched f
Hi,
in the manager app of Tomcat 5.5 there is the list of
threads and their state and connection for each Connector
when you click on "Server Status".
The AJP connector shows a large number (about 50) threads,
all in state "K" (keepalive) with connection times of
several to several hundred hours
Hi Louise,
the key setting is the path to the crystalreportsviewer115
directory: It is set in CRConfig.xml in WEB-INF/classes.
But you have probably done this right in the first place.
Can you copy the exact error message about not finding
any classes? Is it a Tomcat error screen? However, you
n
Hi Rob,
check catalina.out, there should be some message of
log4j complaining if it has been loaded but cannot work.
You may also add -Dlog4.debug to the java command line,
this makes log4j output debug messages on stdout.
If you don't see any output of log4j, then I'd suggest
calling java with
Hi,
it is in the Java Server Pages 2.0 Specification:
(http://jcp.org/aboutJava/communityprocess/final/jsr152/)
"As of JSP 2.0, it is illegal to refer to any classes from the unnamed (a.k.a.
default) package. This may result in a translation error on some containers,
specifically those that run
[EMAIL PROTECTED] wrote:
mainapp
*.do
mainapp
/subapp/*
This allows me to route ./mainapp/subapp/myaction.do correctly, but I
can't access ./mainapp/subapp/images/xyz.gif. What I'd really like to do
> is have the last url-pattern to be "/subap
[EMAIL PROTECTED] wrote:
Georg Sauer-Limbach wrote:
the question is: How do you create the output of
the servlet, that is, with which Writer or OutputStream.
yes you're right: I simply used the output stream.
Never do this if you want to output character data.
(Unless you do the enc
Hi,
the question is: How do you create the output of
the servlet, that is, with which Writer or OutputStream.
If you do this:
public void doGet( HttpServletRequest request,
HttpServletResponse response ) throws IOException {
response.setCharacterEncoding( "UTF-8" );
Writ
save these
options with JCA which I don't know enough about.
Georg
Wayne Gemmell wrote:
On Wed, 28 Feb 2007 12:36:03 +0100
Georg Sauer-Limbach <[EMAIL PROTECTED]> wrote:
If you don't want to deal with HTTP, you should
not use the Servlet API (which is the Java
abstraction of HT
If you don't want to deal with HTTP, you should
not use the Servlet API (which is the Java
abstraction of HTTP) at all. You can do the
indicated code with generic sockets, no need to
mind about Servlets altogether.
Georg
Peter Kennard wrote:
Ok - continuing.
Is it possible to use a "GenricServ
David Delbecq wrote:
No, this is not possible, as far as i know, to stop services without
stopping tomcat. However, using the manager interface of each of your
virtual host, you can stop webapps inside those services. Tomcat will
still listen to corresponding port but will respond with 404 error
I think the problem is that you have servlet-api.jar in the
lib directory of your webapp - this jar is already loaded
for each webapp by Tomcat, it is in the common/lib
directory. So remove this jar from the WEB-INF/lib
directory of all your webapps.
Georg
Asaf Lahav wrote:
*When starting tomc
The admin application is by default protected by security-constraints
which are defined in its web.xml file. In the delivery version, the
admin application required a user with the "admin" role.
You may remove the security constraints (not recommended except
for development purposes) or create a
How can I retrieve the host name (defined in the tomcat server.xml) or some
unique host id in a JSP/Servlet code?
request.getRequestURL() returns the complete URL the client
used. If you have multiple s in your tomcat
installation, then here you see the host name used.
See javax.servlet.http.Ht
Darren Hall wrote:
(I wasn't sure how to configure the ResourceBundle in the code. I copied all
the files the resource bundle seemed to be referring to into my /servlet
directory along with the compiled HelloWorldServlet code, but it still
complained on execution that it couldn't locate the reso
Mike Sabroff wrote:
/simple-servlet
should be
/simple-servlet/*
Does not need to. You can specify paths without
wildcards, which then only match that very url.
The whole story about servlet mappings (which is
quite short actually) is concisely explained in
section 11.2 of the Servlet specifica
Darren,
if I look at your original configuration
SimpleServlet
/simple-servlet
---snip---
then, if I am not overlooking something,
the correct URL to invoke the servlet is
http://localhost/simple-servlet/simple-servlet
because the first "simple-servlet" is the
C
Chris Mooring wrote:
Thanks for the info. I think unless I can set up virtual directories on the
Tomcat HTTP server, I might be forced into using another webserver and
connector (jk - soon to be jk3 ;)).
What do you mean exactly by "virtual directories"?
If it is something like mapping URL pat
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