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

2012-01-05 04:52:21
Hello Roelof,

Actually you should write better subject lines, because this subject line 
doesn't say anything about the issue you're dealing with.

I think you use xslt1.0

An simple option is:

<xsl:for-each select='year'>
          <p class="menu_head"><xsl:value-of select='@value'/></p>
        <xsl:apply-templates select='month'/>
         or just do
        <xsl:for-each select='month'>.....etc......</xsl:for-each>
        
</xsl:for-each>

Op 5 jan 2012, om 11:33 heeft Roelof Wobben het volgende geschreven:



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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