RE: [xsl] only display if subnodes occur more than once

["Michael Kay" <mike@xxxxxxxxxxxx>]

> Anrew Welch wrote:
>  
> > <xsl:template match="sub_a|sub_b|sub_c">
> >   <xsl:variable name="copy"
> > select="boolean(*/following-sibling::*[local-name() = 
> > preceding-sibling::*/local-name()])"/>
> > 
> >   <xsl:if test="$copy">
> >     <xsl:copy-of select="."/>
> >   </xsl:if>
> > </xsl:template>
> 
> <xsl:if test="boolean(*/following-sibling::*[local-name() =
> preceding-sibling::*/local-name()])"> works fine for my needs - thank
> you very much! Could'nt have done it without you!
> 

But it's probably O(n^3) in the number of siblings, so don't try it if
there's a large fanout.

Also the original post asked for all the elem_2's to be output: this code
surely omits the last one?

A 2.0 solution that's likely to be more efficient is

<xsl:for-each-group select="*" group-by="local-name()">
  <xsl:if test="current-group()[2]">
    <xsl:copy-of select="current-group()"/>
  </xsl:if>
</xsl:for-each>

(However, this doesn't preserve the order of the children. The OP didn't say
whether that was a requirement.)

Michael Kay
http://www.saxonica.com/