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  • M4TTHEW 21 posts 61 karma points
    Aug 05, 2009 @ 20:40
    M4TTHEW
    0

    Display number of child pages?

    UI Questions

    I would like to display the number of child pages on a parent page, i don't want any other data from them like page names or content or anything, i just want a macro to count the number of child pages and return a number on the parent page.

    i.e.:

    PAGE A "this page has three child pages"

    -page1

    -page2

    -page3

    PAGE B "this page has one child page"

    -page4

     

    Can anyone point me in a helpful direction?

    TIA!

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  • Peter Dijksterhuis 1442 posts 1722 karma points
    Aug 05, 2009 @ 20:49
    Peter Dijksterhuis
    1

    You could create a xslt-macro. Put something like this in the macro:

    <xsl:value-of select="count($currentPage/node)"/>

    HTH,

    Peter

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  • Sebastiaan Janssen 5045 posts 15477 karma points MVP admin hq
    Aug 05, 2009 @ 20:52
    Sebastiaan Janssen
    100

    In XSLT you can do something like this:

     

    <xsl:for-each select="$currentPage/ancestor-or-self::node [@level=$level]/node [string(data [@alias='umbracoNaviHide']) != '1']">
          <a href="{umbraco.library:NiceUrl(@id)}">
               <xsl:value-of select="@nodeName"/> 
         </a>
                    This page has <xsl:value-of select="count(descendant::node)"/> chilpage(s)
    </xsl:for-each>
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  • Sebastiaan Janssen 5045 posts 15477 karma points MVP admin hq
    Aug 05, 2009 @ 21:07
    Sebastiaan Janssen
    0

    My solution actually lists all the subpages and their subpages, so if you have:

    Page C

    - page5
    - page6
    --page7
    --page8

    Then it will say there is 4 childpages.

    Not sure if you will need this, but it could come in handy!

     

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  • M4TTHEW 21 posts 61 karma points
    Aug 06, 2009 @ 20:55
    M4TTHEW
    0

    Thanks to both of you! a few questions though:

    1) Peter: your example gives me three links to the three subpages, "link link link" not a number of how many pages there are "3". I used the "List subpages from a changeable source" macro as a template and replaced the line here: <xsl:value-of select="@nodeName"/> with your line of code.

    2) Sebastiaan: your example gives me an error- "The variable or parameter 'level' is either not defined or it is out of scope." i copy/pasted your code into a clean macro and replaced the <!-- start writing XSLT --> with your code.

    sorry for the hassle, i appreciate the help dudes!

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  • M4TTHEW 21 posts 61 karma points
    Aug 06, 2009 @ 21:25
    M4TTHEW
    0

    Sebastian: i put this: <xsl:param name="level"/> in the macro and now it doesn't give an error, but it doesn't display anything :(

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  • Sebastiaan Janssen 5045 posts 15477 karma points MVP admin hq
    Aug 06, 2009 @ 21:41
    Sebastiaan Janssen
    0

    Matthew, sorry mate, I forgot to give you this as well: 

    <xsl:variable name="level" select="1"/>
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  • M4TTHEW 21 posts 61 karma points
    Aug 06, 2009 @ 22:28
    M4TTHEW
    0

    Sebastian, thanks! :) here's what i ended up using:

    <xsl:for-each select="umbraco.library:GetXmlNodeById($source)/self::node [string(data [@alias='umbracoNaviHide']) != '1']">
            <a href="{umbraco.library:NiceUrl(@id)}">    
                    </a>
                    <xsl:value-of select="count(descendant::node)"/> sub-pages available
    </xsl:for-each>

    this way i can choose which page i want the number of child pages counted from. awesome!

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