Subject: RE: [xsl] Identical entries in different input documents should appear in the output document only once From: "Meyer, Roland 1. (NSN - DE/Germany - MiniMD)" <roland.1.meyer@xxxxxxx> Date: Mon, 10 Sep 2007 07:47:46 +0200 |
Hi David, thanks for that input. I tried both solutions. At least the code can be reduced :-). In runtime I could not see a difference for my example (but I only measured in units of seconds) - when I have time I might check this with detailed measurements. Roland -----Original Message----- From: ext David Carlisle [mailto:davidc@xxxxxxxxx] Sent: Friday, September 07, 2007 10:55 PM To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx Subject: Re: [xsl] Identical entries in different input documents should appear in the output document only once > That's all ... Wow :-) it's a bit more than you need though:-) <xsl:for-each select="current-group()[1]"> <!-- In my case all blocks is (always) the same thing as <xsl:for-each select="."> <!-- In my case all blocks which is always the same thing as <!-- In my case all blocks so the code can be <xsl:for-each-group select="document($allFiles)//root/block" group-by="idTag"> <!-- In my case all blocks from different files are really identical --> <xsl:call-template name="makeMyTransformationOnTheBlock"/> </xsl:for-each-group> If <root> is always the top level element, using /root/block instead of //root/block would save the processor a lot of work as // implies (except when Michael optimizes it away) a full search to arbitrary depth of each document to find all elements of that name. David ________________________________________________________________________ The Numerical Algorithms Group Ltd is a company registered in England and Wales with company number 1249803. The registered office is: Wilkinson House, Jordan Hill Road, Oxford OX2 8DR, United Kingdom. This e-mail has been scanned for all viruses by Star. The service is powered by MessageLabs. ________________________________________________________________________
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