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Re: [xsl] what's the best way to do this

2012-01-05 08:15:00
i Roelof,

to be complete:

<xsl:variable name="maanden" select="'januari', 'februari', 'maart', 'april', 
'mei', 'juni', 'juli', 'augustus', 'september', 'oktober', 'november', 
'december'"/>

<xsl:template match="year">
        <div id="first-pane" class="menu-list">
                <p class=menu_head"><xsl:value-of select="@value"/></p>
                <div class="menu-body">
                        <xsl:apply-templates select="month"/>
                </div>
        </div>
</xsl:template>

<xsl:template match="month">
        <a href="#"><xsl:value-of select="index-of('maanden', 
number(@value))"/></a>
</xsl:template>

Best regards,

Peter
On Jan 5, 2012, at 11:33 , Roelof Wobben wrote:



Hello, 



Last question and then I have the site ready.



I have now this xml : 



<menu> 

    <section id="9" handle="dagboek">Dagboek</section> 

          <year value="2005"> 

          <month value="03"> 

              <entry id="17" /> 

          </month> 

           <month value="02"> 

               <entry id="16" />

               <entry id="15" /> 

               <entry id="14" /> 

           </month> 

         </year> 

</menu>



And I like to have this output : 



<div id="firstpane" class="menu_list"> 

    <p class="menu_head">2005</p>    

    <div class="menu_body">       

        <a href="#">februari</a>

        <a href="#">maart </a>    

    </div>  

</div>



And I wonder now what's the best approach to get all the months under the 
year ?



Regards,



Roelof

                                        
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