RE: [xsl] Top 100 rows and sort them?

["Michael Kay" <mike@xxxxxxxxxxxx>]

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.Holmberg@xxxxxxxxx] 
> Sent: 05 June 2006 12:54
> To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> 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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