Re: [xsl] Re: [xslt transform & grouping] Using the Muenchian Method?

["Michael PG" <xrow@xxxxxxx>]

Hi Anton,

I combine grouping and filtering, and when I do that I don't get and 
results.
Only thing that new output XML generates is:

<Documents>
</Documents>

I use the solution that you suggested:

  <xsl:template match="Documents">
      <Documents>
          <xsl:for-each 
select="Document[@filter=$filter]/Article[count(.|key('by-info',@info)[1])=1]">
              <Document name="{@info}">
                  <xsl:copy-of 
select="key('by-info',@info)[@filter=$filter]"/>
              </Document>
          </xsl:for-each>
      </Documents>
  </xsl:template>


The input XML looks like:

<Documents>
  <Document chapter="1" title="title 1" href="file1.xml" filter="">
        <Article title="1.1" info="sub" filter="food"/>
        <Article title="1.2" info="main" filter="drink"/>
   </Document>
  <Document chapter="2" title="title 2" href="file2.xml" filter="food">
        <Article title="2.1" info="sub" filter="drink"/>
        <Article title="2.2" info="main" filter="food"/>
   </Document>
</Documents>


thank you.

/Michael


> From: Anton Triest <anton@xxxxxxxx>
> Reply-To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> Subject: Re: [xsl] Re: [xslt transform & grouping] Using the Muenchian 
> Method?
> Date: Thu, 30 Sep 2004 09:00:49 +0200
>
> Michael PG wrote:
>
> > I still get empty parent nodes that are not filtered, since I left out the 
> > info. that attribute filter is also present on the parent node. It can 
> > also be empty or contain information.
> >
> > How can I filter that away as well.
>
> If you say you left out the info, does that mean you don't group anymore, 
> only filter? And in the example below, what do you want to do with Article 
> forename="John"? If you filter out the elements with filter='food', the 
> first Document will not be included. That would leave out John too, 
> although he has a filter='food'.
>
> > <Documents>
> >   <Document id="0001" filter="">
> >       <Article title="Mr"/>
> >       <Article forename="John" filter="food"/>
> >       <Article surname="Smith" filter=""/>
> >   </Document>
> >   <Document id="0002" filter="food">
> >       <Article title="Dr"/>
> >       <Article forename="Amy" filter=""/>
> >       <Article surname="Jones" filter="food"/>
> >   </Document>
> > </Documents>
>
> The best solution depends on what exactly you want to do. If it's plain 
> filtering, a slight variation on the identity template would do:
>
>    <xsl:template match="*">
>        <xsl:copy>
>            <xsl:copy-of select="@*"/>
>            <xsl:apply-templates select="*[@filter='food']"/>
>        </xsl:copy>
>    </xsl:template>
>
> that would give:
>
>    <Documents>
>        <Document id="0002" filter="food">
>            <Article surname="Jones" filter="food"/>
>        </Document>
>    </Documents>
>
> It might be a good idea to put the filter string in a global parameter:
>
>    <xsl:stylesheet ...>
>        <xsl:param name="filter" select="'food'"/>
>        ...
>        <xsl:apply-templates select="*[@filter=$filter]"/>
>
> If you still want to combine grouping and filtering, that's also possible:
>
>    <xsl:template match="Documents">
>        <Documents>
>            <xsl:for-each 
> select="Document[@filter=$filter]/Article[count(.|key('by-info',@info)[1])=1]">
>                <Document name="{@info}">
>                    <xsl:copy-of 
> select="key('by-info',@info)[@filter=$filter]"/>
>                </Document>
>            </xsl:for-each>
>        </Documents>
>    </xsl:template>
>
> Best regards,
> Anton

_________________________________________________________________
Get ready for school! Find articles, homework help and more in the Back to 
School Guide! http://special.msn.com/network/04backtoschool.armx