You can do this quite entirely from XSLT if i understand you correct. I assume level 2 is 2nd tier:

<xsl:for-each select=$currentPage/ancestor-or-self::node [@level = 2 and string(data [@alias = 'ShowLinkToSubpages']) = '1']/node">
<a href="{umbraco.library:NiceUrl(@id)}"><xsl:value-of select="@nodeName" /></a>
</xsl:for-each>

Ops, some bad grammar there ... and the select missed a " in front.

Thanks for your reply--

That code is giving me an error: Error reading XSLT file: \xslt\SidebarContent.xsl.  And I added the leading ".

Anyways, I need to be able to access the parent of This page Specifically, not everything on a certain Level.  If I am ON a certian level, I need to display Children, and if I am another, different certain level, I ned to display Siblings.  Understand?

Thanks again,

Garrett

The Xpath will only match nodes with that property so it wont display children of other nodes. However, if you could do a <textarea><xsl:copy-of select="$currentPage" /></textarea> and copy the XML here it would be easier to help.

Hi Garrett,

Checking for a property on the parent page is done using the parent:: axis:

<xsl:if test="$currentPage/parent::node[data[@alias = 'ShowLinkToSubpages'] = 1]">
    ...
</xsl:if>

- or using the shorthand "..":

<xsl:if test="$currentPage/..[data[@alias = 'ShowLinkToSubpages'] = 1]">
    ...
</xsl:if>

Listing the siblings of $currentPage is done by taking the children of $currentPage's parent. Messing around with preceding-sibling:: and following-sibling:: isn't really feasible here; so:

<xsl:for-each select="$currentPage/../node">
    ...
</xsl:for-each>

- or, if you're like me, using templates instead of for-each:

<xsl:apply-templates select="$currentPage/../node" />

/Chriztian

Rock n roll, Chriztian!  This code is exactly what I needed, works very well.  I appreciate your time :)

//Garrett

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