On Wed, Oct 13, 2010 at 8:23 AM, Scott Trenda
<Scott(_dot_)Trenda(_at_)oati(_dot_)net> wrote:
New XSLT 2.1 feature? :)
Could be something as:
keys(keyname as xs:string) as item()*
and an additional overload:
keys(keyname as xs:string, document as document-node()) as item()*
--
Cheers,
Dimitre Novatchev
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Truly great madness cannot be achieved without significant intelligence.
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To invent, you need a good imagination and a pile of junk
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Never fight an inanimate object
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You've achieved success in your field when you don't know whether what
you're doing is work or play
~ Scott
-----Original Message-----
From: Michael Kay [mailto:mike(_at_)saxonica(_dot_)com]
Sent: Wednesday, October 13, 2010 10:15 AM
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: Re: [xsl] Count items in a key ?
On 13/10/2010 3:23 PM, Fabien Tillier wrote:
Hi List.
I would like to know if it is possible to get all "key" item from an
xsl:key ?
I mean:
when you define a key, say:
<xsl:key name="mykey" match="NODE" use="@id"/>
You can retrieve all NODE nodes with a given @id attribute.
But is it possible to retrieve all distinct @id values from the key ?
(I know it is possible to get them from the XML file, but wouldn't it be
faster to get what is already retrieved ?)
No, this isn't available. I don't think there's any good reason for not
providing it.
Michael Kay
Saxonica
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