Hi,
Can you read my mail again.
In your code,
../parent::node(.) should be name(../parent::node()) though Iam not
comfortable with the construct ../parent::node() yet. But I suppose you
don't have a choice as you may not know the name of the node beforehand in
which case you don't need this test in the first place.
or use DC's suggestion
<xsl:if test="../parent::node() = self::Dynamic1">
....
Note that I have not used single quotes around Dynamic1 as I did earlier as
this is a node comparison as opposed to the string comparison I did
earlier.
Cheers,
Omprakash.V
Rahil
<qamar_rahil(_at_)ya To:
xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
hoo.co.uk> cc: (bcc: omprakash.v/Polaris)
Subject: Re: [xsl] building path
expressions around dynamic element
05/18/2005 node names
05:59 PM
Please respond
to xsl-list
Sorry still some problems.
David Carlisle wrote:
Also, I wanted to know how to access <Dynamic1> when the control is
inside B/Class
lots of ways:
ancestor::Dynamic1
or
../parent::Dynamic1
So I tried Omprakash and your solution to checking a node name as well
as accessing the node from a grand-child into one but there are problems
with my syntax. Here is what I tried
<xsl:if
test="../parent::node(.)=..//preceding-sibling::A/Class/text()"><!
--Checking
if the ancestor node equals the A/Class text. -->
<here>nfhg</here>
</xsl:if>
Tried this as well
<xsl:variable name="AValue" select="..//preceding-sibling::A/Class/text
()"/>
<xsl:if test="../parent::node(.)=$AValue"><!--Checking if the ancestor
node equals the A/Class text. -->
<here>nfhg</here>
</xsl:if>
Am I at a tangent from the right solution and what is it ?
Thanks
Rahil
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