xslt - muenchian grouping

Question

I was wondering how this predicate([1]), is hardcoded as 1 always in the muenchian grouping. The concept was not clear for me, after a lot of search. It is explained as the current node, is compared with the 1st group returned by the key. Why does it always compare with the first one that a key is matched? Also why are we giving contact[count(. | key('contacts-by-surname', surname)[1]) = 1], the =1 part? again 1 is hardcoded. I referred the below link

http://www.jenitennison.com/xslt/grouping/muenchian.html

Answer 1

Let's say we have a key definition <xsl:key name="contacts-by-surname" match="contact" use="surname"/>, then the expression key('contacts-by-surname', 'Doe') gives you a node set with all contact elements where the surname is Doe. The expression key('contacts-by-surname', 'Doe')[1] gives you the first contact in that "group".

Now when processing all contact elements with for-each or apply-templates we usually want a way to identify the first contact element in each group. This can be achieved with <xsl:for-each select="contact[count(. | key('contacts-by-surname', surname)[1]) = 1]"> or <xsl:for-each select="contact[generate-id() = generate-id(key('contacts-by-surname', surname)[1])]">.

If your requirement is different and you for instance wanted to identify the last item in each group then you could of course use a different predicate, as in <xsl:for-each select="contact[count(. | key('contacts-by-surname', surname)[last()]) = 1]"> or <xsl:for-each select="contact[generate-id() = generate-id(key('contacts-by-surname', surname)[last()])]">.