Tomcat 3.1b and Cocoon_1.7.1 question

[Steven Maring <St...@trcinc.com>]

I feel like I am sooooo close now to getting this working.  :-)

My attempts to subscribe to the tomcat-users@jakarta.apache.org are failing
so forgive the crosspost if this question seems a bit Tomcat centric.

The org.apache.tomcat.util.xml.XmlMapper contains some code that looks like
this:

if(System.getProperty("org.xml.sax.parser") != null )

	parser=ParserFactory.makeParser();
else

	parser=ParserFactory.makeParser("com.sun.xml.parser.Parser");



So my attempts to start Tomcat throw an exception since I don't have( and
don't want to use ) XML4J!

How do I specify org.xml.sax.parser as a system property for Tomcat?  And
what value is it looking for?


Steve Maring

Re: Tomcat 3.1b and Cocoon_1.7.1 question

[Mike Engelhart <me...@earthtrip.com>]

Steven Maring wrote:

> parser=ParserFactory.makeParser("com.sun.xml.parser.Parser");
> 
> 
> 
> So my attempts to start Tomcat throw an exception since I don't have( and
> don't want to use ) XML4J!
Tomcat doesn't use XML4J at all.  XML4J was IBM's implementation which is
now Xerces.  You need to make sure the xml.jar file that comes with Tomcat
is in your classpath.  That is Sun's XML parser.  The obove line says
"com.sun.xml" not com.ibm.xml or whatever IBM used.

mike