Re: [xsl] Position of a Child

[Jörg Heinicke <joerg dot heinicke at gmx dot de>]

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.


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