XSL Iterate through list of items - loops

Question

I have an rss feed which is pulling from an abandoned 'Joke of the day' blog. Because the blog is no longer updating, I want to iterate through the list and display a different item each day, ideally in chronological order.
How do I identify the 'first' item in the list (the oldest post) and then show the next item each day?
The rss source is here: http://feeds.feedburner.com/DailyJokes-ACleanJokeEveryday?format=xml
The full list is here: http://dailyjokes.somelifeblog.com/
My XSL code is here, and currently displays the item at position 2, which is the most recent post:
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="html"/>
<xsl:template match="/">
<xsl:apply-templates select="//item[position() < 2]"/>
</xsl:template>
<xsl:template match="item">
<content-item>
<h1><xsl:value-of select="substring-before(title, ' #Joke')"/></h1>
<p><xsl:value-of select="substring-before(description, '<a href')" disable-output-escaping="yes"/></p>
</content-item>
</xsl:template>
</xsl:stylesheet>
I am surfacing this feed in a SharePoint 2013 RSS webpart
My goal is to display a different item each day but I might settle for simple randomization. Also if anybody can suggest a free or for-cost 'Joke of the day' site which is appropriate for my work intranet, that would be appreciated!

Answer 1

My goal is to display a different item each day
To accomplish your goal, your processor needs to know the current date.
The following is a minimized example showing how to use the date in order to retrieve a different item every day:
XML
<rss>
<channel>
<item>
<title>Alpha</title>
</item>
<item>
<title>Bravo</title>
</item>
<item>
<title>Charlie</title>
</item>
<item>
<title>Delta</title>
</item>
<item>
<title>Echo</title>
</item>
</channel>
</rss>
XSLT 1.0
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
<xsl:strip-space elements="*"/>
<xsl:param name="current-datetime">2017-04-27T00:00:00</xsl:param>
<xsl:template match="/rss">
<xsl:variable name="JDN">
<xsl:call-template name="JDN">
<xsl:with-param name="date" select="$current-datetime"/>
</xsl:call-template>
</xsl:variable>
<xsl:variable name="n" select="count(channel/item)" />
<output>
<xsl:apply-templates select="channel/item[$JDN mod $n + 1]"/>
</output>
</xsl:template>
<xsl:template match="item">
<item>
<xsl:value-of select="title"/>
</item>
</xsl:template>
<xsl:template name="JDN">
<xsl:param name="date"/>
<xsl:param name="year" select="substring($date, 1, 4)"/>
<xsl:param name="month" select="substring($date, 6, 2)"/>
<xsl:param name="day" select="substring($date, 9, 2)"/>
<xsl:param name="a" select="floor((14 - $month) div 12)"/>
<xsl:param name="y" select="$year + 4800 - $a"/>
<xsl:param name="m" select="$month + 12*$a - 3"/>
<xsl:value-of select="$day + floor((153*$m + 2) div 5) + 365*$y + floor($y div 4) - floor($y div 100) + floor($y div 400) - 32045" />
</xsl:template>
</xsl:stylesheet>
In this example, the result will be:
<?xml version="1.0" encoding="UTF-8"?>
<output>
<item>Bravo</item>
</output>
Live demo: http://xsltransform.net/pNmBxZK
If your processor supports the EXSLT date:date-time()extension function, you can modify your stylesheet's header to:
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:date="http://exslt.org/dates-and-times"
extension-element-prefixes="date">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
<xsl:strip-space elements="*"/>
<xsl:param name="current-datetime" select="date:date-time()"/>
For SharePoint, I am guessing you would need to use the TodayIso function in the ddwrt namespace instead.