Hi Francesco,
The Spring configuration component [1] was in fact already used by
default when running the applications as a web application, so that that
actually did not solve the issue.
My beans for each application are specified in a XML placed
META-INF/cocoon/spring (globally accessible
Hello,
I'm using cocoon to generate images from SVG using batik's SVG engine.
In the server logs, I get an error message for each image that is generated:
WARN btpool0-1 org.apache.cocoon.serialization.SVGSerializer - Unable to set
document base URI to null, will default to
Yep, do it in the setup
public void setup(SourceResolver resolver, Map objectModel, String src,
Parameters par) throws ProcessingException, SAXException, IOException
{
super.setup(resolver,objectModel,src,par);
try
{
Source inputSource = resolver.resolveURI(super.source);
this.source =
Hi there,
I am using a Cocoon jar (Version 2.11) file compiled on a 32 bit XP in a
64 bit Tomcat (Version: 7.0.22 ) and using Java 64 bit
(jre-6u24-windows-x64)
It runs against a 32 bit Postgresql database (V8.4).
It all seems to work well unless I call some special functionality in
the
On 11 October 2011 17:38, Paul Joseph pjos...@gmail.com wrote:
Hi there,
I am using a Cocoon jar (Version 2.11) file compiled on a 32 bit XP in a 64
bit Tomcat (Version: 7.0.22 ) and using Java 64 bit (jre-6u24-windows-x64)
It runs against a 32 bit Postgresql database (V8.4).
It all seems
Hi Paul,
Cocoon can't be compiled in 32 or 64 bit as such; it's java so is compiled
to platform-independent java byte code. The byte code is run on a jvm, an
architecture-specific application, and if you're using a jvm with a
just-in-time compiler that translates the byte code (or parts of
Hi Andy,
Thank you very much for your crystal clear explanation, I get it now.
Following your suggestion, I traced back the code and find that the
function below is the culprit. I wrote it in an attempt to return the
root directory of the Tomcat application, and it has somehow managed
to
If you run Tomcat from said folder (C:\Program Files\Apache Software
Foundation\apache-tomcat-7.0.22\) all u need is to find the current
working directory.
I think System.out.println(new File(.).getAbsolutePath()) will print
the working directory. It will have the . at the end (at least in
