Re: XPath for the selected node

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Subject: Re: XPath for the selected node
From: Abel Braaksma <abel.online@xxxxxxxxx>
Date: Thu, 19 Jul 2007 03:44:30 +0200

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

Current Thread
XPath for the selected node v vijith - 19 Jul 2007 01:22:22 -0000 Abel Braaksma - 19 Jul 2007 01:44:00 -0000 <= Dimitre Novatchev - 19 Jul 2007 02:06:57 -0000

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XPath for the selected node, v vijith	Thread	Re: XPath for the selected no, Dimitre Novatchev
Re: [xslt 2.0] Local function, Abel Braaksma	Date	XSLT 2.0: Cast does not allo, Sam Byland
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