Hi,
<?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()">
Nope, because "node()" pattern uses the child axis, and while attributes are
nodes, they are accessible only using the "attribute" axis or abbreviation "@"
. Thus use
<xsl:template match="@*|node()">
...
Cheers,
Jarno
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list