From: Geurt Lagemaat (Lagemaat@oriana-automatisering.com)
Date: Thu Jan 30 2003 - 02:59:00 PST
Must be simple but I cant find it.
I have a stylesheet like this:
<?xml version='1.0' encoding='ISO-8859-1'?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:fo="http://www.w3.org/1999/XSL/Format" version="1.0">
<xsl:output method="xml" version="1.0" indent="yes" encoding="ISO-8859-1"/>
<!-- =============================================================== -->
<!-- Publication Params -->
<!-- =============================================================== -->
<xsl:param name="volume" select="'18'" />
<xsl:param name="volume.number" select="'1'" />
<xsl:param name="volume.year" select="'2002'" />
<xsl:param name="initial.page.number" select="1" />
I invoke XEP with a batchfile that uses the supplied transform.bat file:
C:\XEP\transform.bat -xml C:\XEP\Projects\MyProject\source\source.xml -xsl
C:\XEP\Projects\MyProject\Stylesheets\MyStylesheet.xsl -out
C:\XEP\Projects\ MyProject\PDF\MyResult.pdf -format PDF
Now I like to know how to change a stylesheet param by specifying its value
in the batchfile.
Does someone have an example?
Yours,
Geurt Lagemaat
Oriana Automatisering
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This archive was generated by hypermail 2.1.5 : Thu Jan 30 2003 - 02:53:15 PST