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Perfomance Problem

Michael Kay mike at saxonica.com
Mon Feb 22 15:26:34 PST 2010


  Perfomance Problem
Performance always depends on the processor you are using. This query will
be inefficient using a processor (such as Saxon-HE) that does serial search
to implement
 
    $coleccion_cod-agrupa_xi/BLOQUE_CODAGRUPA[AGRUPA1=$distinct_agrupa1
 
and it will be much more efficient on one (such as Saxon-EE) that optimizes
this expression using indexing or hash tables.
 
As an alternative to finding a processor with a suitable optimizer, you
could look for one that implements the XQuery 1.1 "group by" construct, or
an equivalent vendor extension. (Again, Saxon-EE fits the bill).
 
Or you could use XSLT 2.0, which has the convenient <xsl:for-each-group>
construct as standard.
 
Regards,

Michael Kay
http://www.saxonica.com/
http://twitter.com/michaelhkay 


  _____  

From: http://x-query.com/mailman/listinfo/talk [mailto:http://x-query.com/mailman/listinfo/talk] On Behalf
Of Julio de la Vega
Sent: 22 February 2010 14:45
To: http://x-query.com/mailman/listinfo/talk
Subject:  Perfomance Problem



Hi *,

I have problems with performance in the next xQuery

 

 

let $doc := doc("/Lib/doc)

let $res := document {

 

<root>

{

                let $coleccion_cod-agrupa_xi := document 

                {

                               for $XI in $doc/root/XI

                               return

                               (

                               <BLOQUE_CODAGRUPA>

 
<AGRUPA1>{concat($XI/R00000/COD-ENT-GR, $XI/R00000/N-INT-PERS,
$XI/R00000/N-ITN-DIRE )}</AGRUPA1>

 
<AGRUPA2>{concat($XI/R00000/COD-ENT-GR, $XI/R00000/N-ITN-DIRE, "-2"
)}</AGRUPA2>

 
<AGRUPA3>{concat($XI/R00000/COD-ENT-GR, $XI/R00000/N-ITN-DIRE, "-3"
)}</AGRUPA3>

                               </BLOQUE_CODAGRUPA>

                               )

                }

                               return

                                                               

                               for $distinct_agrupa1 in
distinct-values($coleccion_cod-agrupa_xi/BLOQUE_CODAGRUPA/AGRUPA1)

                               

                               return

                

 

                               <TERCIOS>

 
<AGRUPA1>{$distinct_agrupa1}</AGRUPA1>

 
<AGRUPA2>{$coleccion_cod-agrupa_xi/BLOQUE_CODAGRUPA[AGRUPA1=$distinct_agrupa
1][1]/AGRUPA2/text()}</AGRUPA2>

                               </TERCIOS>

                

 

                

 

}

</root>

 

}

return $res/root

 

 

How could I get AGRUPA2 without reading again all the information from
$coleccion_cod-agrupa_xi?

<AGRUPA2>{$coleccion_cod-agrupa_xi/BLOQUE_CODAGRUPA[AGRUPA1=$distinct_agrupa
1][1]/AGRUPA2/text()}</AGRUPA2>

 

 

Any ideas?

 

Thank you

                

 

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