making a plain navigation structure - XSLT

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Filip Lundby 128 posts 149 karma points

Jan 14, 2010 @ 11:15

Making a plain navigation structure?

I would like to change my navgation from this...

<ul>
  <li class="seleted">Top1
    <ul>
      <li>Sub1</li>
      <li class="seleted">Sub2
        <ul>
          <li class="seleted">SubSub1</li>
          <li>SubSub1</li>
          <li>SubSub1</li>
        </ul>
      </li>
      <li>Sub3</li>
    </ul>
  </li>
  <li>Top2</li>
  <li>Top3</li>
</ul>

To this ...

<ul>
  <li class="seleted">Top1</li>
  <li>Top2</li>
  <li>Top3</li>
</ul>
<ul>
  <li>Sub1</li>
  <li class="seleted">Sub2</li>
  <li>Sub3</li>
</ul>
<ul>
  <li class="seleted">SubSub1</li>
  <li>SubSub2</li>
  <li>SubSub3</li>
</ul>

Should be easy, right? But my brain wont work toaday :-S

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Filip Lundby 128 posts 149 karma points

Jan 14, 2010 @ 12:13

0

After a brain-reboot it's running agian :-) Will post solution ...

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Filip Lundby 128 posts 149 karma points

Jan 14, 2010 @ 12:21

This is a simplified version of my xslt ...

<xsl:template match="/">

  <xsl:call-template name="menu">
    <xsl:with-param name="level" select="1"/>
  </xsl:call-template>

</xsl:template>


<xsl:template name="menu">
  <xsl:param name="level"/>
  <xsl:variable name="navNodes" select="$currentPage/ancestor-or-self::node [@level=$level]/node [string(data [@alias='umbracoNaviHide']) != '1']" />

  <ul>
    <xsl:for-each select="$navNodes">
      <li><xsl:value-of select="@nodeName"/></li>
    </xsl:for-each>
  </ul>


  <xsl:for-each select="$navNodes">

    <xsl:if test="$currentPage/ancestor-or-self::node/@id = current()/@id">
      <xsl:call-template name="menu">
        <xsl:with-param name="level" select="$level+1"/>
      </xsl:call-template>
    </xsl:if>

  </xsl:for-each>

</xsl:template>

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