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RE: [xsl] Sort by calculated value

2006-08-29 06:55:18
Your second sort key count(//Outward[Id=Id]/SegmentList/Segment) has the
same value for all the nodes in the node-set being sorted: there are two
things obviously wrong with it, the initial "//" which selects from the
root, and the predicate [Id=Id] which will always be true. Perhaps you
intended  

count(//Outward[Id=current()/Id]/SegmentList/Segment)

but I can't tell without seeing the source.

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

-----Original Message-----
From: Giancarlo Rossi [mailto:giancarlo_rossi(_at_)tiscali(_dot_)it] 
Sent: 29 August 2006 14:20
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [xsl] Sort by calculated value 


I'm trying to sort using a calculated value Process MSXML Xsl 1.0


For. E.g

<xsl:for-each
select="/CommandList/CheckRouting/RouterList/Router/GroupList/
Group/OutwardL
ist/Outward" > <xsl:sort select="Price/Amount" data-type="number"
order="ascending" /> <xsl:sort
select="count(//Outward[Id=Id]/SegmentList/Segment)" 
data-type="number"
order="ascending" />

'my code
</xsl:for-each>

But the second sort is not respected.

Infact the "segment" could be 1 or 2..

Id is the Id of the outward

http://www.lastminutesud.it/test/last_mad_lgw.xml

I hope you could help me.




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