CMS umbraco-XSL后代

Question

I have a page with a folder and links inside

-FirstPage
  +folder
  -page
  -page
  -folder
    -link1
    -link2
    -link3
  -page

I want to reach the link1,link2,link3 in the tree. My current page is FirstPage. How do I do that?? This is the xsl i wrote and he give me the first link in the top folder

<xsl:template match="/">
<xsl:for-each select="$currentPage/descendant-or-self::* [@isDoc][@level=2]">
<xsl:if test="count(current()/descendant::* [@isDoc]) &gt; 0">
<xsl:variable name="descendantPage" select="current()/descendant::* [@isDoc]"/>   
<xsl:value-of select="$descendantPage/text"/>
</xsl:if>
</xsl:for-each>
</xsl:template>

Thank you for your help.

edit:new xsl I use...

<xsl:variable name="fId" select="number(1395)" />
<xsl:variable name="linksFolder" select="$currentPage/descendant-or-self::* [@isDoc][@level=2][@id='$fId']">
<xsl:template match="/">
  <xsl:for-each select="$linksFolder/* [@isDoc]">
    <xsl:value-of select="./text">
  </xsl:for-each>
</xsl:template>

How can i avoid to use a output (like the id of the folder) to get the folder i want to? Thanks for your help...

Answer 1

There are lots of helpful tips & tricks in the Umbraco WikiBook online ( http://en.wikibooks.org/wiki/Umbraco/Various_useful_XSLT_methods_in_Umbraco ). You can use the doc type of the folder node to find the links (instead of the id).

For example looping round all nodes of a certain doc type (linkFolderDocType):

<xsl:for-each select="$currentPage/ancestor-or-self::root//node [@nodeTypeAlias='linkFolderDocType']">
  <xsl:value-of select="./text">
</xsl:for-each>