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  • Steve 472 posts 1216 karma points
    Jun 06, 2014 @ 22:17
    Steve
    0

    Listing Multiple Nested Node Names on Select

    Hi,

    I have some news articles that are stored by a node "category" and children "year". Using xslt, I am pulling in the parent node of the article, which is the year the article was published, but I would also like to pull in the category of the article as well, but I don't quite know how to go about it. Here is the section my existing xslt that is pulling the parent node of the article.

    <div class="featured-articles">
      <xsl:for-each select="$featuredArticles">
        <div class="listing">
          <p class="tag">
            <xsl:call-template name="node-name">
              <!-- can't do ./parent::* because these nodes are copied and don't have parents -->
        <xsl:with-param name="node" select="$currentPage/ancestor-or-self::NewsHome//NewsArticle[@id=current()/@id]/parent::*" />
            </xsl:call-template>
          </p>
          
    <!-- A template to output the correct name for a given node. Better than copy/pasting this code all over the place -->
<xsl:template name="node-name">
  <xsl:param name="node" />

  <xsl:if test="string($node) != ''">
    <xsl:choose>
      <xsl:when test="$node/pageTitle != ''">
        <xsl:value-of select="$node/pageTitle"/>
      </xsl:when>
      <xsl:otherwise>
        <xsl:value-of select="$node/@nodeName"/>
      </xsl:otherwise>
    </xsl:choose>
  </xsl:if>
</xsl:template>
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  • Kim Nedergaard 37 posts 144 karma points
    Jun 07, 2014 @ 13:47
    Kim Nedergaard
    100

    Hi Steve,

    Just to make sure that I understand you correctly. Your sitetree looks like this:

    • Newshome
      • Category
        • Year
          • Article

    and you want to your xslt to output: Category - Year - Articlename?

    Then you could do the following:

    First find your articlenode:

    <xsl:variable name="sourceNode" select="$currentPage/ancestor-or-self::NewsHome//NewsArticle[@id=current()/@id]"/>

    Then call you template for the three nodes and put the output in variables to keep your markup clean:

        <!-- First category -->
<xsl:variable name="categoryName">
         <xsl:call-template name="node-name">              
            <xsl:with-param name="node" select="$currentPage/ancestor-or-self::NewsHome//NewsArticle[@id=current()/@id]/ancestor::category" />
         </xsl:call-template></xsl:variable>

    <!-- Then year -->
<xsl:variable name="yearName">
        <xsl:call-template name="node-name">              
                <xsl:with-param name="node" select="$currentPage/ancestor-or-self::NewsHome//NewsArticle[@id=current()/@id]/ancestor::year" />
         </xsl:call-template></xsl:variable>

    <!-- Finally name -->
<xsl:variable name="articleName">
        <xsl:call-template name="node-name">              
                <xsl:with-param name="node" select="$currentPage/ancestor-or-self::NewsHome//NewsArticle[@id=current()/@id]" />
         </xsl:call-template>
</xsl:variable>

<!-- And then output your results -->
<xsl:value-of select="$categoryNAme"/> - <xsl:value-of select="$yearName"/> - <xsl:value-of select="$articleName"/>

    Hop this answers your question :)

    -Kim

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  • Steve 472 posts 1216 karma points
    Jun 09, 2014 @ 15:18
    Steve
    0

    Works great Kim, but I don't understand why we even have to declare the variable "sourceNode" since we never refrence it? Could you help me understand?

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  • Kim Nedergaard 37 posts 144 karma points
    Jun 10, 2014 @ 08:49
    Kim Nedergaard
    0

    Sorry, I were a bit quick on the copy+paste there :)

    Instead of calling: 

    <xsl:call-template name="node-name">              
                <xsl:with-param name="node" select="$currentPage/ancestor-or-self::NewsHome//NewsArticle[@id=current()/@id]/ancestor::year" />
</xsl:call-template>

    You could call: 

     <xsl:call-template name="node-name">              
                <xsl:with-param name="node" select="$sourceNode/ancestor::year" />
 </xsl:call-template></xsl:variable>
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