Re: How to sort?

How do I get Only one chapter e.g. chapter="2" instead of getting all chapters
in kombination with implementation above?

-Mike


On Fri, 15 Oct 2004 10:07:34 +0200, Mike G <row.filter@xxxxxxxxx> wrote:
> 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
> >
> >
> 
> 
> --
> 
> [row.filter]
> 


-- 

[row.filter]