Re: [xsl] Testing the name of parent node

Subject: Re: [xsl] Testing the name of parent node
From: Joerg Heinicke <joerg.heinicke@xxxxxx>
Date: Fri, 20 Dec 2002 11:19:15 +0100
Hello Ragulf,

Ragulf Pickaxe wrote:
Hello all,

I think this is trivial, but I can't seem to find how to do it:

I am in a node <a...> which can have any number of parents or childs of the same type. I want to know if the node I have in hand is the topmost <a...>.

the solution:


<xsl:if test="not(ancestor::a)">
  no a above me in the tree
</xsl:if>

<xsl:if test="not(ancestor::*[name() = 'a'])">
  the same here
</xsl:if>

I tried looking at the name() funktion, but the explanation was not very clear to me (Michael Kay's book, 2nd edition), and I don't know if this is the right track at all.

the explanation:


name() returns the complete name of an element. So a <a/> in input returns the string 'a', <html:a/> returns 'html:a' and so on.

I really would appreciate some pointers.

Thank you
Ragulf Pickaxe :)

Regards,


Joerg


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