[XSL-LIST Mailing List Archive Home]
[By Thread]
[By Date]
Hi Spencer,
Many thanks to you and to everybody else who posted hints or complete solutions to my problem!
Re: [xsl] eliminating duplicates
Subject: Re: [xsl] eliminating duplicates From: Garvin Riensche <g.riensche@xxxxxxx> Date: Thu, 10 May 2007 23:30:17 +0200 |
Hi Spencer,
Many thanks to you and to everybody else who posted hints or complete solutions to my problem!
Regards, Garvin
Hi Garvin,
if your using xslt 2.0 it's pretty easy:
<xsl:for-each-group select="//edge" group-by="@source"> <xsl:for-each-group select="current-group()" group-by="@target"> <xsl:copy> <xsl:copy-of select="@*"/> </xsl:copy> </xsl:for-each-group> </xsl:for-each-group>
if your using 1.0 you may have to wait for someone with more knowledge of keys.
Spencer
On 5/10/07, Steve <subsume@xxxxxxxxx> wrote:On 5/10/07, Garvin Riensche <g.riensche@xxxxxxx> wrote: > If there was only one attribute, lets say "source" it would be simple: > <xsl:for-each > select="//edge[not(./@source=preceding-sibling::edge/@source)]"> > <xsl:copy-of select="."/> > </xsl:for-each>
That wouldn't really work either. That would just assure you didn't spit out a duplicate of the preceding record.
Someone will surely followup with more comprehensive suggestions about xsl:key. I'd research it until then. =)
-Steve
Current Thread |
---|
|
<- Previous | Index | Next -> |
---|---|---|
Re: [xsl] eliminating duplicates, Spencer Tickner | Thread | Re: [xsl] eliminating duplicates, Abel Braaksma |
RE: [xsl] eliminating duplicates, cknell | Date | Re: [xsl] "re-calling" a template f, bryan rasmussen |
Month |
Keywords