How do I select the last N elements with XPath?

Question

I support a web site which generates content XML that is then translated into web pages using XSLT. I have been asked to create a new stylesheet which will transform the ou

Answer 1

position()/last() returns position/last position within the current context, so when the navigator is positioned in one <month>, position() will return <day> within that month, and last() will return last <day> within that month, but i guess you know that.

Therefore, what you could do is flatten all <day>'s in an array and put in a variable, prior to selecting just like you did before.

<xsl:variable name="days" select="//day"/>
<xsl:apply-templates select="$days[position()>last()-30]" />

Answer 2

Browsing through the XSLT spec today, I found a note which explains why // behaves this way:

// is short for /descendant-or-self::node()/. For example, //para is short for /descendant-or-self::node()/child::para and so will select any para element in the document (even a para element that is a document element will be selected by //para since the document element node is a child of the root node); div//para is short for div/descendant-or-self::node()/child::para and so will select all para descendants of div children.

NOTE: The location path //para[1] does not mean the same as the location path /descendant::para[1]. The latter selects the first descendant para element; the former selects all descendant para elements that are the first para children of their parents.

In other words, when using //, the position() is calculated along the child axis, not the descendant-or-self axis. Specifying descendant or descendant-or-self allows you to get the first/last n nodes as you'd expect:

<xsl:apply-templates select="descendant::day[position()>last()-30]"/>