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

[Boudewijn Bosgoed <bbosgoed@xxxxxxx>]

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