Re: [xsl] Display the number of records in a group

Subject: Re: [xsl] Display the number of records in a group
From: anil kumar <anar.leti@xxxxxxxxx>
Date: Tue, 9 Mar 2010 11:59:23 +0530
Hi,

Sorry if i confused you.

Let me explain more clearly

My xml consists of 11 records ,
================================
123 ABC GOOD 10
123 ABC GOOD 20
123 ABC GOOD 30

456 DEF GOOD 10
789 GHI GOOD 10
567 JKL GOOD 10

234 BCD GOOD 20
345 CDE GOOD 20

678 TTT TRIANGULATION 30
678 TTT TRIANGULATION 20

321 TRT TRIANGULATION 20
================================

when i used  group-by="concat(VAT_TAX_NO,BILL_CUST_NAME,INTRA_EU_TRX_CODE)">
These 11 records are grouping  into "8" records.
and i want to capture this value and show it in o/p

these are the ways i tried..
<xsl:value-of select="count(current-group())"/>   giving o/p as
<mezo eazon="0A0001D003A">31111121</mezo>
<xsl:value-of select="last()"/>                          giving o/p as
  <mezo eazon="0A0001D003A">1111111111111111</mezo>
<xsl:value-of select="position()"/>                    giving o/p as
<mezo eazon="0A0001D003A">12345678</mezo>


expected o/p
<mezo eazon="0A0001D003A">8</mezo>

Thanks
Anil

On Mon, Mar 8, 2010 at 6:46 PM, David Carlisle <davidc@xxxxxxxxx> wrote:
> On 08/03/2010 13:04, anil kumar wrote:
>>
>> How to show the toatal number of records,
>> ima failing to get this by  last()
>> because iam  grouping 11 records into 8 , so the value of last() is
>> giving 11  which is in correct
>>
>>
>
> I didn't understand that comment, but inside xsl:for-each-group
>
> <xsl:value-of select="last()"/> gives you the number of groups
>
> <xsl:value-of select="count(current-group())"/> gives the number of items
in
> the current group.
>
> I'm not sure which of these numbers you want.
>
> David
>
>
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