RE: [xsl] Listing things in columns

[Jarno.Elovirta@xxxxxxxxx]

Hi,

> Each Crate may have from 1 to more than 100 Oranges. When I list them
> alphabetically, in paragraph form, the code looks like this:
> 
> <xsl:for-each
> select="//Object[@id=$CrateID]/Relationships/Relationship[Name
> ='Has_Orange']
> /ValueList/Value/Reference">
> 	<xsl:sort select="." data-type="text" order="ascending"/>
> 	<xsl:variable name="OrangeName" select="."/>
> 	<xsl:variable name="OrangeID" select="./@Id"/>
> 	<a href="{$OrangeID}"><xsl:value-of select="$OrangeName"/></a>	
> 	<xsl:if test="position() != last()">, </xsl:if>
> </xsl:for-each>
> 
> But this is messy. Instead, I would like to list them in 
> three columns side
> by side. 
> 
> I tried the following:
> <table>
>  <tr>
>   <xsl:for-each
> select="//Object[@id=$CrateID]/Relationships/Relationship[Name
> ='Has_Orange']
> /ValueList/Value/Reference">
>     <xsl:variable name="OrangeName" select="."/>
>     <xsl:variable name="OrangeID" select="./@Id"/>
>     <td>
> 	<a href="{$OrangeID}"><xsl:value-of select="$OrangeName"/></a>
>     </td>
>     <xsl:if test="position() mod 3 = 0"> [something here to 
> end a row and
> start a new row]</xsl:if>
>   </xsl:for-each>
>  </tr>
> </table>
> 
> But I couldn't make anything work inside the if test, because 
> it doesn't
> like to see </tr><tr>.

Yours is a grouping problem--see <http://jenitennison.com/xslt/grouping>. If you want to sort the References, you have to use exslt:node-set() or two passes.

Cheers,

Jarno - Assemblage 23: Breath Of Ghosts (143 BPM)

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