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RE: [xsl] XML Question

From: "Michael Kay" <mike@------------>
To:
Date: 10/4/2006 11:23:00 AM
The name() function takes a single node as its argument, not a list of
nodes.

Change name(ancestor::*) to ancestor::*/name()

Michael Kay
http://www.saxonica.com/
 

> -----Original Message-----
> From: Per Osnes [mailto:per.osnes@xxxxxx] 
> Sent: 04 October 2006 12:18
> To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> Subject: SV: [xsl] XML Question
> 
> Hi folks,
> This template gives the required output (in Altova).
> 
> <xsl:template match="//*[not(*)]">
> <xsl:variable name="path">
> <xsl:for-each select="ancestor::*"><xsl:value-of
> select="name()"/>/</xsl:for-each>
> <xsl:value-of select="name()"/>
> </xsl:variable>
> <xsl:text>
> </xsl:text>
> <xsl:value-of select="$path"/>
> <xsl:value-of select="' ===> '"/>
> <xsl:value-of select="text()"/>
> </xsl:template>
> 
> Actually I wanted to replace the "for each" by something like
>   <xsl:value-of select="string-join(((name(ancestor::*)), 
> name()), '/')"/>
> 
> but Altova gives error message "too many items"...
> I have used this method earlier for producing the full path 
> for elements in an XML schema:
>   select="string-join((ancestor::xs:element/@name, ./@name), '.')"
> 
> Any comments from the experts?
> 
>  - Per Osnes.
> 
> 
> ----- LINKE Markus <markus.linke@xxxxxxxx> wrote:
> > Hi,
> > 
> > I am not sure if this is the right list for such a 
> question, but lets 
> > try :)
> > 
> > I would like to create a list of xpaths via xsl from a given 
> > xml-structure like this:
> > 
> > <a>
> >    <b>
> >       <c>
> >          content
> >       </c>
> >       <d>
> >          morecontent
> >       </d>
> >    </b>
> > </a>
> > 
> > it should show the full xpath like
> > 
> > a/b/c  ====>   content
> > a/b/d  ====>   morecontent
> > 
> > and not just
> > 
> > c => content
> > d => morecontent
> > 
> > Ideas anyone? I am looking for either a script or some xsl ...
> > 
> > Cheers,
> > Markus


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