Re: [xsl] XPath for the selected node

[Abel Braaksma <abel.online@xxxxxxxxx>]

I don't know much about xquery, but in XSLT you can do this on any path:

<xsl:value-of select="ancestor-or-self::*/name()" separator="/" />

Or, not tested, the following if you need the depth as well:

<xsl:value-of  separator="/">
   <xsl:for-each select="ancestor-or-self::*">
      <xsl:value-of select="name()" />
      <xsl:text>[</xsl:text>
      <xsl:value-of select="count(preceding-sibling::*[name() = 
current()/name()]" />
      <xsl:text>]</xsl:text>
   </xsl:for-each>
</xsl:value-of>


Cheers,
-- Abel

v vijith wrote:
> Dear All,
> I have an XML which has the structure like
>
> <root>
>     <data>
>          <node1>Welcome </node1>
>          <node2 refid="22">Some data</node2>
>      </data>
>      <anotherdata>
>               <sometype refid=33>welcome</sometype>
>       </anotherdata>
> </root>
>
> Using Xquery is there a way to find the xpath of all the nodes that
> has refid as an attribute? For the above case, the answer that I would
> be looking at it
>
> root/data/node2
> root/anotherdata/sometype
>
> I went through some of the posts and found that saxon:path is an
> option. Do I have any other way of doing it?
>
> Any kind of pointer is appreciated!!
>
> Thanks
> Vij