Re: We need to kick someone out of the group

Hi all,

Sometimes I am baffled about how to approach something. Sometimes I'm

Using XSLT 2.0, I have the input XML:

     <contrib-group>
         <contrib><name>Bob</name><xref rid="aff1"/></contrib>
         <contrib><name>Judy</name><xref rid="aff2"/></contrib>
     </contrib-group>
     <aff id="1"><label>1</label>Kingdom of Curds</aff>
     <aff id="2"><label>2</label>Land of Whey</aff>
     <contrib-group>
         <contrib><name>Jimmy</name><xref rid="aff3"/></contrib>
     </contrib-group>
     <aff id="3"><label>2</label>Duchy of Lambic-Soaked Cheese Rind</aff>

I'm trying to get:

     <contrib-group>
         <contrib><name>Bob</name><xref rid="aff1"/></contrib>
         <contrib><name>Judy</name><xref rid="aff2"/></contrib>
         <aff id="1">Kingdom of Curds</aff>
         <aff id="2">Land of Whey</aff>
     </contrib-group>
     <contrib-group>
         <contrib><name>Jimmy</name><xref rid="aff3"/></contrib>
         <aff id="3">Duchy of Lambic-Soaked Cheese Rind</aff>
     </contrib-group>

That is, move the following <aff> elements into the <contrib-group> and

I would use group-starting-with in a template matching the parent or
node e.g.

   <xsl:template match="*[contrib-group]">
        <xsl:copy>
            <xsl:for-each-group select="*"
group-starting-with="contrib-group">
                    <xsl:copy>
                        <xsl:apply-templates select="@*, node(),
tail(current-group())"/>
                    </xsl:copy>
            </xsl:for-each-group>
        </xsl:copy>
    </xsl:template>

Assumes the identity template set up. tail(current-group()) is
subsequence(current-group(), 2) in XSLT 2.