Re: [xsl] Re: Sorting on two levels

["Martin Honnen martin.honnen@xxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>]

Michele R Combs mrrothen@xxxxxxx wrote:
> The suggestion offered yesterday works great as far as the sorting, but I need to do certain things to each item in a group as it's output, such as wrapping each ID in an <a> element.  What I tried is shown below, but it results in @href containing *all* the ids for the group.  How do I process each item in a group separately?  Do I need to at some point just do <xsl:apply-templates select=" current-group()"> and then rely on templates for primary, secondary, etc.?  I did try that but couldn't seem to get it to work -- either I got no output, or I got the entire group in a chunk.

>                              <a>
>
>                                 <xsl:attribute name="href">
>
>                                      <xsl:value-of select="current-group()/@id"/>
>
>                                 </xsl:attribute>
>
>                                 <xsl:value-of select="current-group()/@id" separator=", "/>

If you want to map each item in a group to an HTML "a" element then use

  <xsl:for-each select="current-group()">
    <a href="{@id}">
      <!-- not sure what you want to output for the link -->
      <xsl:value-of select="@id"/>
    </a>
  </xsl:for-each>

or use

  <xsl:apply-templates select="current-group()" mode="link"/>

and write a template

  <xsl:template match="indexterm" mode="link">
    <a href="{@id}">
      <!-- not sure what you want to output for the link -->
      <xsl:value-of select="@id"/>
    </a>
   </xsl:template>