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Re: [xsl] matching elements of a list
Subject: Re: [xsl] matching elements of a list From: Rolf Schumacher <mailinglist@xxxxxxxxx> Date: Tue, 08 May 2007 23:37:57 +0200 |
Thank you very much, Jeff. Worked right away on that sample. Targeting it to my more complex problem (XMI spec) is not as easy because the links are not just a sibling. However with your example I should be able to figure out how that will work. Thank you, again Rolf Jeff Sese wrote: > try this stylesheet, i think this would work (if i guessed your > requirements right). > > <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" > version="1.0"> > <xsl:template match="/"> > <xsl:apply-templates select="root/st"/> > </xsl:template> > <xsl:template match="st"> > <xsl:value-of select="concat(@ix,':')"/> > <xsl:apply-templates > select="following-sibling::ln[@s=current()/@ix]"/> > <xsl:text>
</xsl:text> > </xsl:template> > <xsl:template match="ln"> > <xsl:value-of select="concat(' ', @d)"/> > <xsl:apply-templates select="following-sibling::ln[@s=current()/@d]"/> > </xsl:template> > </xsl:stylesheet> > > -- > Jeff > > > Rolf Schumacher wrote: >> do I have reached the limits of xpath? >> >> if several linked lists are contained in one document how to match all >> nodes belonging to a specific start node? >> >> The following example may illustrate my question: >> >> Input: >> <root> >> <st ix="a"/> >> <st ix="b"/> >> <el ix="c"/> >> <el ix="d"/> >> <el ix="e"/> >> <el ix="f"/> >> <el ix="g"/> >> >> <ln s="a" d="g"/> >> <ln s="g" d="f"/> >> >> <ln s="b" d="e"/> >> <ln s="e" d="c"/> >> <ln s="c" d="d"/> >> </root> >> >> disired output: >> >> a: g f >> b: e c d >> >> How to accomplish that by XSLT (2.0)? >> Even your answer "I know for sure that there is not elegant way" >> or "You got to extend the processor by some Java function" >> would help.
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