Hello,
I am trying to transform an xml file into another xml file by means of a
stylesheet. This stylesheet takes as input parameter the location of the DTD
belonging to the generated xml file. What I would like to do is:
<xsl:param name="DTDLocation"/>
<xsl:output method="xml" encoding="UTF-8"
doctype-system="{$DTDLocation}"/>
This is possible in XSLT 1.1, but not in XSLT 1.0. Another way is to write
an extension function that writes the DOCTYPE to the output. However I
prefer a solution in XSLT 1.0 (if I am correct XSLT 1.1 became never a
Recommendation) and without extension functions. Is this possible somehow ?
Thanks in advance.
Kind regards,
Ismaël
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list