Re: [xsl] How to sort?

[Mike G <row.filter@xxxxxxxxx>]

Yeah, this works with some modification.

 <xsl:apply-templates select="Document[Article/@info = 'main' or
Article/@filter = 'food']"/>

thanx!

-Mike


On Thu, 14 Oct 2004 15:06:16 +0300, jarno.elovirta@xxxxxxxxx
<jarno.elovirta@xxxxxxxxx> wrote:
> Hi,
> 
> > What I do now, is choose all Article elements that contain info="main"
> > and info="sub" attributes and then sort everything by state="n"
> > attribute.
> 
> An Article cannot have two info attributes, thus the above condition can never be true in XML.
> 
> > Is it possible to implement an sorting option that enables Article
> > elements where filter="food" to be placed on the top of the Article
> > list, that is giving them higher priority in sort than the previous
> > value of state="n".
> >
> > so the XML output will look like:
> >
> > <?xml version="1.0"?>
> > <Documents>
> >       <Document title="1" chapter="i" href="file1.xml" filter="food">
> >               <Article title="1.2" info="main" filter="food"/>
> >               <Article title="1.3" info="main" filter="drink"
> > state="2"/>
> >       </Document>
> >
> >       <Document title="2" chapter="ii" href="file2.xml"
> > filter="drink">
> >                      <Article title="2.2" info="main" filter="food"/>
> >                      <Article title="2.1" info="main"
> > filter="drink" state="1"/>
> >       </Document>
> >
> >       <Document title="4" chapter="2" href="file2.xml" filter="">
> >               <Article title="3.2" info="main" filter="food"/>
> >       </Document>
> > </Documents>
> 
>  <xsl:template match="Documents">
>    <xsl:copy>
>      <xsl:apply-templates select="Document[Article/@info = 'main' and Article/@info = 'sub']"/>
>    </xsl:copy>
>  </xsl:template>
>  <xsl:template match="Document">
>    <xsl:copy>
>      <xsl:copy-of select="@*"/>
>      <xsl:for-each select="Article[@info = 'main']">
>        <xsl:sort select="@filter ='food'" data-type="number" order="descending"/>
>        <xsl:sort select="@state" data-type="number"/>
>        <xsl:copy-of select="."/>
>      </xsl:for-each>
>    </xsl:copy>
>  </xsl:template>
> 
> I'm not sure exactly how you wanted the info attribute handled, because you desired output doesn't use Articles with "sub" info. Anyhow, this produces the desired output. The
> 
>  <xsl:sort select="@filter ='food'" data-type="number" order="descending"/>
> 
> sort condition basically works so that the "@filter = 'food'" expression will either return boolean true or false, which will be cast to number 1 or 0, respectively, and that gives you the sort key value.
> 
> 
> 
> Cheers,
> 
> Jarno - Lisa Lashes: Hard Mix
> 
> 


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