Re: New twist: eliminating nodes with duplicate content, cas

David Carlisle wrote:
> 
> > The code below does it in one pass (it was indeed a brave assertion!),
> 
> I suspect that was multiple passes by Michael's counting (as is the code
> below) I don't think that you can do it in a single xpath expression,
> for the reasons given, but I don't think you need to use explicit
> recursion. Isn't this more or less what is wanted?

I thought Michael meant it couldn't be done without piping the source
through 2 stylesheets. - Anyway, your solution is much more readable
than mine, and more efficient at least until XSL processors are smart
enough to optimize that recursion into iteration - which I assume they
aren't yet.

> <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
>                 version="1.0"
>                 >
> 
> <xsl:variable name="up" select="'ABCDEFGHIJKLMNOPQRSTUVWXYZ'"/>
> <xsl:variable name="lo" select="'abcdefghijklmnopqrstuvwxyz'"/>
> 
> <xsl:template match="/">
> <xsl:apply-templates select="//handle"/>
> </xsl:template>
> 
> <xsl:template match="handle">
> <xsl:if test="not(following::handle[translate(.,$up,$lo)=
>                                     translate(current(),$up,$lo)])">
>   <xsl:copy-of select="."/>
> </xsl:if>
> </xsl:template>
> 
> </xsl:stylesheet>
> 
> Of course the above probably results in the current node being downcased
> multiple times and it would be better anyway not to use
> current() at all and just put the lowercase of the current node value
> into a variable.

-- 

bahhumbug

phil

*witness relocation program alumnus*


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