Doesn't the use of position() suffice to have the item number?
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Diese Nachricht wurde von meinem Android Mobiltelefon mit GMX Mail
gesendet.Am 18.08.20, 16:09 schrieb "Terry Ofner tdofner@xxxxxxxxx"
<xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>:
Is it possible to reference elements in p[@class=bnlb] (to the
left of the ! $groups) in the iterate? Or to pass a parameter
including the item number of the p[@class=bnlb] to the iterate.
The selected node of the iterate is <p=class=bDirectionsb>.
On Aug 13, 2020, at 3:55 PM, Martin Honnen martin.honnen@xxxxxx <xsl-list-service@lists.
mulberrytech.com> wrote:
<xsl:template match="set[p[@class = 'nl']]">
<xsl:variable name="groups" as="map(xs:string, element())*">
<xsl:for-each-group select="p[@class = 'directions']/*"
group-starting-with="span[@class = 'letter']">
<xsl:sequence select="map { 'letter' : ., 'term' :
current-group()[2] }"/>
</xsl:for-each-group>
</xsl:variable>
<write_choices>
<xsl:iterate select="p[@class = 'nl'] ! $groups">
<write_choice num="{position() - 1}" letter="{?letter}"
term="{?term}"/>
</xsl:iterate>
</write_choices>
</xsl:template>
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