[xep-support] Who to use XSLT with Java-implementation

[Bastian Kroll]

Hi,
I´ve got a problem wih using XSLT-Stylesheets.
First my szenario:
I have a XML-File, from this file I call the XSL-Stylesheet. When I´m using
the command-line with
java -jar /<path_to_the_xep>/xep.jar -xml /<path_to_the_xml>/Pageheader.xml
It´s running well without errors.

Now I have a Java-Programm and want XEP to use the stylesheet as well. The
call is:
public ClassMaker _owner  = null;    // owner
com.renderx.xep.FormatterImpl _xep = new
com.renderx.xep.FormatterImpl(_optionsXEP);
XEPRenderer renderer = new XEPRenderer(_config,_owner);
renderer.renderTemplate(pdfdummy,xml_file,_xep);

But now I´m getting a Exception
com.renderx.xep.lib.FormatterException: Invalid XSL FO source: 1 error found
during validation
    at 
com.renderx.xep.FormatterCore$ContentHandlerDriver.endDocument(Unknown
Source)
    at com.icl.saxon.aelfred.SAXDriver.parse(SAXDriver.java:334)
    at com.renderx.xep.FormatterCore.render(Unknown Source)
    at com.renderx.xep.FormatterImpl.render(Unknown Source)
    at com.renderx.xep.FormatterImpl.render(Unknown Source)
    at com.classwizard.cw.XEPRenderer.renderTemplate(XEPRenderer.java:61)
    at com.classwizard.cw.PageHeader.replaceText(PageHeader.java:367)
    at 
com.classwizard.cw.FillMultiCreateThread.run(FillMultiCreateThread.java:118)


I´ve got no Iidea, how to make XEP to use the XSL Transformation. Can
anybody give me some help?
In the attachment is my XML- an XSL-File.

I´m using Java 1.5 and XEP 4.5 build 20060313


mit freundlichen Grüssen / best regards

Bastian Kroll
ClassWizard GmbH
Heidenkampsweg 45
D-20097 Hamburg

Tel. +49 40 80 90 45-133
Mobil +49 172 45 11 474

mailto:bk at classwizard.com
http://www.classwizard.com

ClassWizard GmbH ist Eingetragen beim Registergericht Eutin unter HRB 880

Geschäftsführende Gesellschafter Heribert Hanisch und Jörg Schiessler

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