You need two phases: first sort by number, then by word. That means you
either need XSLT 2.0, or the XSLT 1.0 xx:node-set() extension, or two
stylesheets running in a pipeline.
In 2.0:
<xsl:variable name="sorted-by-number" as="element(row)*">
<xsl:perform-sort select="/root/row">
<xsl:sort select="xs:integer(number)" order="descending"/>
</xsl:perform-sort"/>
</xsl:variable>
<xsl:variable name="top-100" select="$sorted-by-number[position() le 100]"/>
<xsl:for-each select="$top-100">
<xsl:sort select="word"/>
<xsl:copy-of select="."/>
</xsl:for-each>
Michael Kay
http://www.saxonica.com/
-----Original Message-----
From: Niklas Holmberg [mailto:Niklas(_dot_)Holmberg(_at_)eniro(_dot_)com]
Sent: 05 June 2006 12:54
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [xsl] Top 100 rows and sort them?
hi!
This is my xml (contains about 400 row elements):
<root>
<row>
<word>Afrikanskt</word>
<number>21</number>
</row>
<row>
<word>Amerikanskt</word>
<number>23</number>
</row>
</root>
what i want to do is to pick the 100 elements that has the
largest "number" and then print them out sorted by "word". is
this possible to do at all with xsl?
thanks!
/niklas
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