Edward Wohlman wrote:
> Is it possible to use xsl to perform a topological sort such that data 
> such as
> 
>     <Route>
>         <Place>B</Place>
>         <Place>D</Place>
>         <Place>F</Place>
>     </Route>
>     <Route>
>         <Place>A</Place>
>         <Place>B</Place>
>         <Place>C</Place>
>         <Place>E</Place>
>     </Route>
>     <Route>
>         <Place>C</Place>
>         <Place>D</Place>
>         <Place>E</Place>
>     </Route>
>     <Route>
>         <Place>A</Place>
>         <Place>E</Place>
>         <Place>F</Place>
>     </Route>
> 
> can be transformed to
> 
>    <Route>
>        <Place>A</Place>
>        <Place>B</Place>
>        <Place>C</Place>
>        <Place>D</Place>
>        <Place>E</Place>
>        <Place>F</Place>
>    </Route>

The simply looks like an alphabetical sort of distinct Place contents, 
like this:

<xsl:stylesheet
   xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
   version="1.0">

<xsl:strip-space elements="*"/>

<xsl:output method="xml" indent="yes"/>

<xsl:key name="distinct-place" match="Place" use="."/>

<xsl:template match="/*">
   <xsl:copy>
     <Route>
       <xsl:apply-templates select="Route/Place[generate-id() = 
generate-id(key('distinct-place', .)[1])]">
         <xsl:sort select="." data-type="text"/>
       </xsl:apply-templates>
     </Route>
   </xsl:copy>
</xsl:template>

<xsl:template match="Place">
   <xsl:copy>
     <xsl:apply-templates/>
   </xsl:copy>
</xsl:template>

</xsl:stylesheet>


-- 

	Martin Honnen --- MVP XML
	http://JavaScript.FAQTs.com/