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Re: [xsl] generate full xpath name to an attribute?


Subject: Re: [xsl] generate full xpath name to an attribute?
From: Ray Tayek <rtayek@xxxxxxxxx>
Date: Mon, 31 Mar 2003 06:43:20 -0800

At 02:57 PM 3/31/03 +0200, you wrote:
Ray Tayek wrote:
hi, i need to generate the full xpath name (location) for attributes in the following xsl (see below).
instead of using <xsl:attribute name="name"><xsl:value-of select="name()"/></xsl:attribute> which gets me the just the attribute (field) name, i would like to generate the full xpath expression from the root of the document (i guess it would look something like: /......./Node/SingleForces@attribute_name). i.e. what do i use in instead of: <xsl:value-of select="name()"/> to get the fully qualified name for this field from the root.

See faq: http://www.dpawson.co.uk/xsl/sect2/N6077.html -- Oleg Tkachenko http://www.tkachenko.com/blog Multiconn Technologies, Israel

yikes! - looks like the simplest way is to have:


<xsl:template match="name">
       <xsl:for-each select="ancestor-of-self::*">
         <xsl:value-of select="name()" /><xsl:text>/</xsl:text>
       </xsl:for-each>
       <xsl:value-of select="." />
     </xsl:template>

in *every* atribute of the document. i suppose one could build up a variable that had the path to the current node, but that sees like an awful lot of work. i will try this out though.

one woul think that the parser would just know this and could emit it in a canonical form. hard to believe that there is no function call to do this.

thanks for your prompt assistance.

---
ray tayek http://tayek.com/ actively seeking mentoring or telecommuting work
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