Sign in Register
Copied to clipboard

Flag this post as spam?

This post will be reported to the moderators as potential spam to be looked at


  • Jon 10 posts 31 karma points
    Apr 01, 2010 @ 13:23
    Jon
    0

    Don't render empty ul

    What

    Don't render empty ul

    Why

    1. empty ul is invalid (x)html => validating page with w3 validator to fail
    2. CSS set on menu (for ex: different background color/image ) renders "clutter"

    Solution

    Proposed solution:

    surround ul element with conditional

    <xsl:if test="$parentNode/node or $calculatedMednuDepth = 1 and $forceHome">

    <ul>
    ...

     

    Means:

    Only render ul-menu if there are children ;
    with exception if forceHome is selected and it's the 1'st level (the actual home link)

    Copy Link
  • Claire Botman 48 posts 77 karma points
    Jan 10, 2011 @ 02:22
    Claire Botman
    1

    It looks like this suggestion was put into the current code, but we still get empty ul if there are docs that have umbracoNaviHide set, so I changed the test to this:

    <xsl:if test="$parentNode/*[@isDoc][string(umbracoNaviHide) != '1'] or ($calculatedMenuDepth = 1 and $forceHome)">
    Copy Link
  • Johan 188 posts 380 karma points
    Nov 09, 2011 @ 19:40
    Johan
    0

    Thanks Claire! This should really be included. For me it messed up a lot more than just the semantics. The "isBranch" variable also needs this condition (if you use it). Like this:

    <xsl:variable name="isBranch">
  <xsl:choose>
    <xsl:when test="$currentPage/ancestor-or-self::*[@isDoc][@id = $currentNodeID]/child::*[@isDoc][string(umbracoNaviHide) != '1'] and string(count(descendant::umbracoNaviHide[text() = '0'])) != '0'">1</xsl:when>
  </xsl:choose>
</xsl:variable>
    Copy Link

  • This forum is in read-only mode while we transition to the new forum.

    You can continue this topic on the new forum by tapping the "Continue discussion" link below.

Please Sign in or register to post replies