Subject: Re: [xsl] Count items in a key ? From: Dimitre Novatchev <dnovatchev@xxxxxxxxx> Date: Wed, 13 Oct 2010 10:50:44 -0700 |
On Wed, Oct 13, 2010 at 8:23 AM, Scott Trenda <Scott.Trenda@xxxxxxxx> 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 --------------------------------------- Truly great madness cannot be achieved without significant intelligence. --------------------------------------- To invent, you need a good imagination and a pile of junk ------------------------------------- Never fight an inanimate object ------------------------------------- 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@xxxxxxxxxxxx] > Sent: Wednesday, October 13, 2010 10:15 AM > To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx > Subject: Re: [xsl] Count items in a key ? > > B 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: >> B B B <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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