Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

How do I select the last N elements with XPath?

Tags:

xml

xslt

xpath

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 output of the "archive" page into Atom for syndication. The problem I'm running into is that the archive page contains a rather large number of items — 142 and counting — and the feed should never have more than thirty items.

Currently, the output from the archive page looks something like this:

<archive>
    <year>
        <month>
            <day>
            <day>
            ...
        </month>

        ...
    </year>

    ...
</archive>

The year and month tags are used by the HTML transform but are completely irrelevant for an Atom feed. I had hoped that using the position() function with the descendant axis would work (//day[position()>last()-30]), but this selects the last 30 days of each month, which isn't at all what I need. :-)

Is there a way to do this with XSLT or XPath? Having to modify the XML generator to add, say, a feed="true" attribute to the last thirty days seems like a pretty nasty kludge.

like image 487
Ben Blank Avatar asked Jan 19 '09 18:01

Ben Blank


1 Answers

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]" />
like image 68
baretta Avatar answered Oct 22 '22 21:10

baretta