I have a document type that is used in different areas of the website. On this document type is a tag group named 'default'.

I have seen other forum posts where someone wants to list pages that contain a given tag. But what I want to do is pull a sub-list of tags based on their location under a given node.

So for example, I might have the following content tree structure:

-site root --sub 1 ---sub 1.1 ---sub 1.2 --sub 2 --- sub 2.1 --- sub 2.2 --- sub 2.3

Note that sub 1.x and sub 2.x use the same document type. I want to generate and display a list of tags on sub 2 which are in use on pages under sub 2 only.

I currently get my tag list like this:

  <xsl:for-each select="tags:getAllTagsInGroup('default')/tags/tag">

I see that there is an xslt extension for tag type: getTagsFromNode(int nodeid)

But what I want is something like getTagsUnderNode (int nodeid).

I realize that another solution would be to just have separate document types where a different tag group could be defined. But I'm trying to avoid that.

Is there some XSLT magic to do it using a predicate?

Actually, I realize, that I do not need to use the extension methods for tags. I can just pull a list of nodes where the tag field is not null, like this:

<xsl:variable name="thetags" select="umbraco.library:GetXmlNodeById($currentPage/@id)//story[tags != '']" />

and then loop through them...

<xsl:for-each select="$thetags">
   <xsl:sort select="." />
    <a href="/taglist?tag={./tags}">
      <xsl:value-of select="./tags"/>
    </a>
</xsl:for-each>

But, I quickly realize, that this is not going to work. I need to somehow create a distinct list, and take into account that some nodes will have 1 tag, and others will have more than one, separated by commas....