Subject: Re: [xsl] Position of a Child From: Jörg Heinicke <joerg.heinicke@xxxxxx> Date: Thu, 20 Sep 2001 20:55:05 +0200 |
I don't know whether there is a shorter version, but the following code will work: <xsl:for-each select="*"> <xsl:if test="name()=$NodeName"> <xsl:value-of select="position()"/> </xsl:if> </xsl:for-each> Joerg > Hi all, > > I'm having problems getting the child's position() of the context node. > I reference the child by its name (stored in a variable), like this: > > <xsl:value-of select="*[name()=$NodeName]"/> > > What I would like to get is the position of that child inside the context > node. > I don't seem to find the correct syntax for this. > > Thanks in advance, > Frank. XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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