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XSL: Docbook to foreign DTD
- From: Jochen Hein <jochen at jochen dot org>
- To: docbook-apps at lists dot oasis-open dot org
- Date: Thu, 03 Jan 2002 21:16:46 +0100
- Subject: DOCBOOK-APPS: XSL: Docbook to foreign DTD
I'm converting Docbook into a given DTD and have problems mapping
xref. It looks pretty easy:
Docbook: <xref linkend="something"/> is going to be
<xref xlink:href="something"/> in the foreign DTD.
My template
<xsl:template match="xref">
<xref>
<xsl:attribute name="xlink:href"><xsl:value-of select="@linkend"/></xsl:attribute>
</xref>
</xsl:template>
generates <xref href="something"/>, so I assume that gives me problems
with some sort of namespace. So my question is, how do I generate an
attribute name with a colon? I don't think I'll have luck in changing
the foreign DTD...
Jochen
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