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Re: [xsl] ordering nodes


Subject: Re: [xsl] ordering nodes
From: andrew welch <andrew.j.welch@xxxxxxxxx>
Date: Mon, 30 Jan 2006 09:51:27 +0000

On 1/29/06, bokluk <bokluk@xxxxxxx> wrote:
> LIST.XML:
>
> <section label="ligostart">
>         <section uri="erzaehlen.html" label="erzaehlen">
>                 <section uri="dererzaehltext.html" label="dererzaehltext">
>                         <section
> uri="dertextalssprachlicheszeichensystemdiscours.html"/>
>                 </section>
>         </section>
> </section>
>
> dererzaehltext.XML:
>
>
> ....
> <definition title="Der Erzdhltext" label="dererzaehltextdef">
>                         <LMMLtext>blablabla</LMMLtext>
> </definition>
> <definition title="Der Erzdhltext2" label="dererzaehltext2def">
>                         <LMMLtext>blabla</LMMLtext>
> </definition>
> .....
>
> Other XML's mentioned in LIST.XML have the same structure as above.
>
> I need to extract this definitions out of every file and create one html
> with all of them ordered alphabetically.
>
> P.S. I'm using cocoon 2 processor

I'm not familiar with the cocoon 2 processor, but if it's an XSLT 2.0
processor you can use:

<xsl:for-each select="for $x in
collection('pathToXMLDirectory?select=*.xml;recurse=yes;on-error=warn')
			return $x//defnition">
  <xsl:sort select="youdidntsay"/>

...where pathToXMLDirectory is the path of the directory containing
the XML and youdidntsay is the value to use in the sort :)

Sorry the answer is a bit vague, but the question is a bit vague too.


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