RE: Generating variable DOCTYPE

Like this

<?xml version="1.0"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
        <xsl:param name="DTDLocation"/>
        <xsl:template match="/">
                <xsl:text disable-output-escaping="yes"><![CDATA[
<!DOCTYPE svg PUBLIC "-//W3C//DTD SVG 20001102//EN" "]]></xsl:text>
                <xsl:value-of select="$DTDLocation"/>
                <xsl:text 
disable-output-escaping="yes"><![CDATA[">]]></xsl:text>
        </xsl:template>
</xsl:stylesheet>

Edward Middleton

-----Original Message-----
From: Cams Ismael [mailto:Ismael(_dot_)Cams(_at_)siemens(_dot_)atea(_dot_)be]
Sent: Thursday, January 16, 2003 5:35 PM
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [xsl] Generating variable DOCTYPE


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





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