xsl-list
[Top] [All Lists]

Re: [xsl] Re: Sorting on two levels

2015-04-15 09:03:54
Michele R Combs mrrothen(_at_)syr(_dot_)edu 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>
--~----------------------------------------------------------------
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
EasyUnsubscribe: http://lists.mulberrytech.com/unsub/xsl-list/1167547
or by email: xsl-list-unsub(_at_)lists(_dot_)mulberrytech(_dot_)com
--~--

<Prev in Thread] Current Thread [Next in Thread>