Subject: Re: New twist: eliminating nodes with duplicate content, case-insensi tive From: David Carlisle <davidc@xxxxxxxxx> Date: Tue, 7 Dec 1999 10:25:49 GMT |
> 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? <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> To do this directly in xpath you would need a generalisation of current() that gave the node current outside the current filter rather than the node current at the start of the expression. 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. David XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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