Hi all,
my very small Stylsheet:
<?xml version="1.0" encoding="iso-8859-1"?>
<xsl:stylesheet
version="1.0"xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" encoding="iso-8859-1" indent="yes"/>
<xsl:template match="node()">
<xsl:copy>
<xsl:apply-templates select="@*"/>
<xsl:apply-templates/>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>
It doesnt copy attribute nodes and i dont know why it doesnt. It
matches the default attribute template which i thought would be
overriden by <xsl:template match="node()">
In my XML Book "In a nutshell" it says:
"node() matches all nodes regardless of type: attribute,
namespace..."
and later in the same book about xsl:template it says:
"templates witch match patterns that just have a node test (*,@*,
node()...) have priority -0.5" like built-in templates, but "they
[built-in templates] are also imported before all other templates,
and thus never override any explicit template rule, regardless of
priority"
so all attributes are selected by <xsl:apply-templates select="@*"/>
and because they match node() and <xsl:template match="node()"> has
same priority as built-in <xsl:template match="text()|@*"> BUT it
has higher import precedence.
So it should match and xsl:copy with attribute should create an
attribute with same name and value as the source document (Kay, XSLT
2nd edition p. 193)
But it doesnt work with xsltproc neither with saxon
Can anyone push me in the right direction to solve my problem (or how
to read my books :-)
regards,
janning
--
Planwerk 6 /websolutions
Herzogstraße 86
40215 Düsseldorf
fon 0211-6015919
fax 0211-6015917
http://www.planwerk6.de
-------------------------------------------------------
--
Planwerk 6 /websolutions
Herzogstraße 86
40215 Düsseldorf
fon 0211-6015919
fax 0211-6015917
http://www.planwerk6.de
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list