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Re: [xsl] Re: Sorting on two levels


Subject: Re: [xsl] Re: Sorting on two levels
From: "Martin Honnen martin.honnen@xxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>
Date: Wed, 15 Apr 2015 14:03:36 -0000

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>


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