RE: [xsl] Identical entries in different input documents should appear in the output document only once

["Meyer, Roland 1. (NSN - DE/Germany - MiniMD)" <roland.1.meyer@xxxxxxx>]

Hi,

thanks again for posting. I just solved the problem by using XSLT 2.0
for-each-group as Mike proposed. It's really great how powerful it is.
This is my really simple code snip:

<xsl:for-each-group select="document($allFiles)//root/block"
group-by="idTag">
   <xsl:for-each select="current-group()[1]"> <!-- In my case all blocks
from different files are really identical -->
      <xsl:call-template name="makeMyTransformationOnTheBlock"/>
   </xsl:for-each>
</xsl:for-each-group>

That's all ... Wow :-)

Roland




-----Original Message-----
From: ext Abel Braaksma [mailto:abel.online@xxxxxxxxx]
Sent: Friday, September 07, 2007 10:28 PM
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Subject: Re: [xsl] Identical entries in different input documents should
appear in the output document only once

G. Ken Holman wrote:
> At 2007-09-07 17:54 +0200, Meyer, Roland 1. (NSN - DE/Germany -
> MiniMD) wrote:
>> I have the following problem. I have a couple of XML documents to
merge
>> to one output document.
>> Each document has the same structure like this:
>> [..........].
>>
>> Is there any other and simpler way to - let's say - memorize the
already
>> written blocks resp. identifiers?
>
> Mike is correct that the variable-based grouping method would work
> with this, but it is far slower than an XSLT 2 approach.  When using
> XSLT 1.0 I find variable-based grouping acceptable for sub-document
> and multi-document grouping.
>
> The code would work along the lines of the following, and it assumes
> that there is an XML structure $files with the list of all the file
> names:
>
>   <xsl:variable name="items"
>                 select="document($files/file/@uri,.)/root/block"/>
>   <xsl:for-each select="$items">
>      <!--walk through all, doing work at first of each unique idTag-->
>      <xsl:if test="generate-id(.)=
>                    generate-id($items[idTag=current()/idTag])">
>         <!--the following executes once for each unique idTag
>             value across all the files-->
>      </xsl:if>
>   </xsl:for-each>
>

I didn't see your (Ken's) post earlier, I find it interesting to compare

this approach to mine. Perhaps this works quite faster than my node-set
+ load-all-documents-at-once method (but here, too, all documents are
loaded at once). The difference is, that in my approach, the varying
document root note is removed (by using template match + copy-of and
then node-set the root node for the copied documents becomes the same
root node for all), which makes it possible to do normal muenchian
grouping, instead of look-up grouping (with variables, I mean).

I hope that the added speed by using a key outweighs the overhead for
using node-set. Also, using copy-of is an atomic operation and generally

performs faster than applying the templates again (but of course, that
is only useful when you do not need additional processing on the
blocks).

Cheers,
-- Abel Braaksma