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
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