Subject: [xsl] RE: >Subject: RE: [xsl] node comparision for equality: Message-ID: <448F3BD6.605 From: "uppaluri satyaprasad" <uppaluris@xxxxxxxxxxx> Date: Tue, 20 Jun 2006 01:09:52 -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.
Current Thread |
---|
|
<- Previous | Index | Next -> |
---|---|---|
RE: [xsl] Recursively merge (aggreg, Michael Kay | Thread | [xsl] Concatenating multiple variab, Mohsen Saboorian |
Re: [xsl] Selecting all nodes betwe, Mukul Gandhi | Date | Re: [xsl] Selecting all nodes betwe, Duncan Anker |
Month |