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Re: How to sort?

Subject: Re: How to sort?
From: Mike G <row.filter@xxxxxxxxx>
Date: Thu, 14 Oct 2004 11:17:44 +0200
xsl sort xsl copy
ok.

If this is the XML with previously missing root-elements

<Documents>
       <Document title="1" chapter="i" href="file1.xml" filter="food">
               <Article title="1.1" info="sub" filter="drink" state="1"/>
               <Article title="1.2" info="main" filter="food" state="3"/>
			   <Article title="1.3" info="main" filter="drink" state="2"/>
       </Document>
       <Document title="2" chapter="ii" href="file2.xml" filter="drink">
               <Article title="2.1" info="main" filter="drink" state="1"/>
               <Article title="2.1" info="sub" filter="drink" state="3"/>
               <Article title="2.2" info="main" filter="food" state="2"/>
       </Document>
       <Document title="3" chapter="1" href="file2.xml" filter="">
               <Article title="3.1" info="sub" filter="drink" state=""/>
               <Article title="3.2" info="child" filter="" state="8"/>
       </Document>
       <Document title="4" chapter="2" href="file2.xml" filter="">
               <Article title="3.1" info="sub" filter="drink" state=""/>
               <Article title="3.2" info="main" filter="food" state="1"/>
       </Document>
</Documents>


could this be correct XSLT for my output. Could you check it? It runs
very slow for some reason.

<xsl:template match="/">
<Documents>
   <xsl:apply-templates select="Documents/Document[Article/@info = 'main']"/>
</Documents>
 </xsl:template>
 <xsl:template match="Document">
   <xsl:copy>
     <xsl:copy-of select="@*"/>
     <xsl:for-each select="Article[@info = 'main']">
       <xsl:sort select="@state" data-type="number"/>
       <xsl:copy-of select="."/>
     </xsl:for-each>
   </xsl:copy>
 </xsl:template>

thanx.

Mike


On Thu, 14 Oct 2004 11:45:38 +0300, jarno.elovirta@xxxxxxxxx
<jarno.elovirta@xxxxxxxxx> wrote:
> Hi,
> 
> > I want pick only Article elements where attribute info="main"
> >
> > info-attribute is not located on the Document level, only on
> > Article level,
> > so when info="main" located it should be placed in the same
> > originating Document node
> > wheter info-atribute is located on Document level or not.
> >
> > I need to order output XML by Chapter, chapter-attribute, on
> > Document-level. (i,ii,1,2 ... n)
> >
> > Then I need to sort the output on Article level by state, ascending.
> >
> > the output XML should look like this
> >
> >
> > <?xml version="1.0"?>
> >
> > <Document title="1" chapter="i" href="file1.xml" filter="food">
> >       <Article title="1.3" info="main" filter="drink" state="2"/>
> >       <Article title="1.2" info="main" filter="food" state="3"/>
> > </Document>
> >
> > <Document title="2" chapter="ii" href="file2.xml" filter="drink">
> >     <Article title="2.1" info="main" filter="drink" state="1"/>
> >     <Article title="2.2" info="main" filter="food" state="2"/>
> > </Document>
> >
> > <Document title="4" chapter="2" href="file2.xml" filter="">
> >       <Article title="3.2" info="main" filter="food" state="1"/>
> > </Document>
> 
>  <xsl:template match="/">
>    <xsl:apply-templates select="Documents/Document[Article/@info = 'main']"/>
>  </xsl:template>
>  <xsl:template match="Document">
>    <xsl:copy>
>      <xsl:copy-of select="@*"/>
>      <xsl:for-each select="Article[@info = 'main']">
>        <xsl:sort select="@state" data-type="number"/>
>        <xsl:copy-of select="."/>
>      </xsl:for-each>
>    </xsl:copy>
>  </xsl:template>
> 
> Note that your output is not well-formed XML and the stylesheet above will recreate the output example.
> 
> Cheers,
> 
> Jarno - Seabound: Contact (Assemblage 23 remix)
> 
> 


-- 

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