Subject: Re: [xsl] Node Position & Relationship! From: António Mota <amsmota@xxxxxxxxx> Date: Fri, 11 Feb 2005 15:24:08 +0000 |
I'm doing it like this: <xsl:variable name="level" select="count(ancestor::Menu)"/> (...) <xsl:choose> <xsl:when test="$level=0"> <xsl:text>1</xsl:text> ----> this node is visible </xsl:when> where Menu is your tree_node. On Sat, 12 Feb 2005 00:08:36 +1000, Adam J Knight <adam@xxxxxxxxxxxxxxxxx> wrote: > Given the following xml structure, I want to create an xsl if element that > tests the current node to see if it is a level 0 tree_node element. > > <?xml version="1.0"?> > <tree> > <tree_node id="7" value="Test"> > <tree_node id="8" value="Test Sub"/> > <tree_node id="9" value="Test Sub One"> > <tree_node id="10" value="Test Sub Two"/> > </tree_node> > </tree_node> > </tree> > > Here is my attached, pretty sad!!!!! > > <xsl:if test="{count(self::*)=1}"> > <xsl:apply-templates select="tree_node"/> > </xsl:if > > Can someone give me the correct way to achieve this. > Also how can I found out if a particular node is a child, ancestor or peer > Of any given node. > > Thanks to anyone who response, muchly appreciated. > > Cheers, > Adam > > NB: "Pray as if everything depended upon God and work as if everything > depended upon man."
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