Subject: RE: [xsl] XSLT Sort - returning attributes From: David Neary <David@xxxxxxxxx> Date: Fri, 16 May 2003 13:17:51 +0200 |
> Basically from the above example, " <xsl:apply-templates > select="*"/> " will > return me all the elements of the book node, however I also > want to retrieve > all attributes. > I 've tried " <xsl:apply-templates select="@*"/> " to get all > attributes, but obviously Im doing something wrong. > Is there any simpler way of going about sorting an xml doc > and returning > the entire document in XML format (only sorted) ? The identity transformation is... <xsl:template match="node() | @*"> <xsl:copy> <xsl:apply-templates/> </xsl:copy> </xsl:template> You could do a template for the root something like this... <xsl:template match="/"> <xsl:apply-templates select="node() | @*"> <xsl:sort select="node whose value you want to use in sort"/> </xsl:apply-templates> </xsl:template> Add in the identity template and you're away in a hack. Cheers, Dave. XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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