Is it possible to group the whole xml document by the Customers node without
hardcoding a specific customer?
and at the same time also use the following date comparison. This code below is
inside a for loop which
lists every <Trade> , so no grouping is introduced. One section of the xml is
listed, but there can be
many Customers and Trade nodes
<xsl:value-of select="format-number(sum(Trade/Step
[concat(substring(MinFlowDate,7),substring(MinFlowDate,1,2),substring(MinFlowDate,4,2))>=
concat(substring($global,7)+3,substring($global,1,2),substring($global,4,2))]
[concat(substring(MinFlowDate,7),substring(MinFlowDate,1,2),substring(MinFlowDate,4,2))<
concat(substring($global,7)+6,substring($global,1,2),substring($global,4,2))]
/StepCharge_TTBlack), '###,###,##0')"/>
</b></td>
structure of xml
<Portfilio>
<AsOf>11/30/2004</AsOf>
<Trade>
<TradeId>ED.TEST</TradeId>
<Customer>AAA</Customer>
<Step>
<MinFlowDate>11/30/2004</MinFlowDate>
<StepCharge_TTBlack>10</StepCharge_TTBlack>
</Step>
<Step>
<MinFlowDate>11/11/2005</MinFlowDate>
<StepCharge_TTBlack>10</StepCharge_TTBlack>
</Step>
</Trade>
<Portfilio>
Thanks,
Dmitri
-----Original Message-----
From: David Carlisle <davidc(_at_)nag(_dot_)co(_dot_)uk>
Sent: Dec 6, 2004 9:42 AM
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: Re: [xsl] grouping and sum
how is it possible to group by the Customer Node to retrieve a sum from
StepCharge?
the sum of all StepCharge for customer AAA is
<xsl:value-of select="sum(Trade[Customer='AAA']/StepCharge)"/>
David
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