Hi,
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)
Yes.
and without extension functions. Is this
possible somehow ?
There is no XSLT 1.0 solution; use your XSLT engine's API to control the
serialization.
Cheers,
Jarno - De/Vision: Endlose Traume
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list