Hi Kristian,

It could be something similar to:

<xsl:for-each select="$currentPage//node [@nodeTypeAlias = 'forumPostAlias']">
  <xsl:if test="position &lt; $numberOfItems">
    <xsl:value-of select="@nodeName"/>
  </xsl:if>
</xsl:for-each>

Key is to use the '//' selector (which selects all children - regardless of their level) instead of '/'. Of course, you'll have to take care of setting starting page (used $currentPage) and declaring the variable numberOfItems and set it to whatever number you'd like, and replacing 'forumPostAlias' with alias of a forum post.

Hope this helps.

Regards,

/Dirk

You also might have a look at this post: http://our.umbraco.org/forum/developers/xslt/3292-How-to-display-the-latest-Blog-Post-(blog-4-umbraco)-on-homepage-of-website?sort=karma Sebastiaan's solutions works for me...

Forum posts are kept in a database, and not in the node structure, so the above won't work for uForum.

I think there is an XSLT extension for getting all posts. Take a look there. 

Hi I am trying to access the latest forum posts/updates using the xslt extension - uForum:LatestTopics, however I am having problems accessing the variables through xslt. This is what i have so far:

<xsl:variable name="utopics" select="uForum:LatestTopics('10')"/>
  <xsl:for-each select="$utopics/topics">
     <xsl:value-of select="./urlname" disable-output-escaping="yes"/>
     <a title="" href="{umbraco.library:NiceUrl(@parentid)}"><xsl:value-of select="./urlname"/></a>
 </xsl:for-each>

This is a real stab at what I am hoping to achieve, but is unfortunately not working. Any guider or pointers would be greatly appreciated

thanks

Andrew

I think you need to do this:

<xsl:for-each select="$utopics/topics/topic">

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