Re: [xsl] position() of parent node

["john prieur" <john2000@xxxxxxxxxxxx>]

Hi,
I would like to try my hand at answering your question (I'm attempting to
stretch a little beyond my true expertise)...

I believe the problem would be that your position() will want to pertain to
the 21st <chapter> child of <imageLink> nodes, but there are no such
children of <imageLink>.
I think you should use a parent::  in the predicate.

----- Original Message -----
From: "Robert Ogden" <Robert.Ogden@xxxxxxxx>
To: <xsl-list@xxxxxxxxxxxxxxxxxxxxxx>
Sent: Monday, November 17, 2003 3:47 PM
Subject: [xsl] position() of parent node


> This should be an easy one, but it is eluding me.
>
> XML:
> <manual>
> <chapter>
> <textLink>para one</textLink>
> <textLink>para two</textLink>
> <imageLink>image one</imageLink>
> <imageLink>image two</imageLink>
> </chapter>
> </manual>
>
> XSLT:
> <xsl:for-each select="chapter">
> <xsl:for-each select="textLink">
> //do some stuff
> </xsl:for-each>
> <xsl:for-each select="imageLink">
> <xsl:if test="chapter[position() != 21]">
> //do your thing
> </xsl:if>
> </xsl:for-each>
> </xsl:for-each>
>
> What I am getting at, is that I want to output all text links (easy
> enough), and all images for chapters unless the chapter is 21 (which
> happens also to be last, which I tried <xsl:if test="chapter[position()
> != last()]">)
>
> Position has given me troubles in the past.
>
> Robert Ogden
> IETM Developer
> Navy Programs
> (763) 572-7121
>
>  XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list
>


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