[xsl] RE: Subject: RE: [xsl] node comparision for equality

Subject: [xsl] RE: Subject: RE: [xsl] node comparision for equality
From: "uppaluri satyaprasad" <uppaluris@xxxxxxxxxxx>
Date: Sun, 25 Jun 2006 15:08:02 -0400
uppaluri satyaprasad wrote:
however i need to also test if
BusPartnerCodeValue (i.e. D06001 here) and BusPartnerCodeTypeCode (i.e. 00003) are equal in two nodes within BPPLCode.
if these two values are equal, the BPPLCode nodes are equal and i shall process only one node.
we can ignore ProductLevel/ProductLevelCode here.

This looks like the "select distinct" problem, which you can solve
with an approach described here:
 http://jenitennison.com/xslt/grouping/index.xml
Use grouping by content, using
concat(BusPartnerCodeType/BusPartnerCodeTypeCode,':',BusPartnerCodeValue)
as key indicator.
(unfortunately, http://jenitennison.com/xslt/unique.xml is a stub).

J.Pietschmann




Hi Pietschmann, Thank you for your help.

I managed to get my problem fixed using  Muenchian method as advised
by you.

I have one more related question. i want to find no of those
distinct nodes (same business rule as above)
when I try to use the keys it will give me the count of the nodes in
the entire document.
however i need to find the count of the unique nodes inside each
BPPLCode.

thanks in advance

Satya

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