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:  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

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