Subject: Re: [xsl] XSL "Joining" From: "Vasu Chakkera" <vasucv@xxxxxxxxxxx> Date: Thu, 02 May 2002 15:30:42 +0000 |
Constraints:
We have to use xmlns:xsl="http://www.w3.org/TR/WD-xsl", so I cannot use any features of the new standard.
its an easy job if u use the namespace declaration <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0"> regards vasu
From: "Cooper, John" <JCooper@xxxxxxxxxxxxxxxxx> Reply-To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx To: <XSL-List@xxxxxxxxxxxxxxxxxxxxxx> Subject: [xsl] XSL "Joining" Date: Thu, 2 May 2002 09:04:10 -0400
Hello, I have the following problem I need help with. I would like to perform something like a join on two different nodes so that I may get a description field from another node. Following is an example of the XML:
XML: <response> <detail @id='1' @parent_key='2' @desc='First Row'/> <detail @id='2' @parent_key='3' @desc='Second Row'/> <detail @id='3' @parent_key='' @desc='Third Row'/> </response>
Problem Statement:
In the xsl, I do a template match on //detail and output each attribute in a table. What is the query to get the @desc attribute by joining the @parent_key attribute to the @id attribute?
Desired Outupt:
ID Name Parent 1 First Row Second Row 2 Second Row Third Row 3 Third Row -
Constraints:
We have to use xmlns:xsl="http://www.w3.org/TR/WD-xsl", so I cannot use any features of the new standard.
Any help would be deeply appreciated.
John Cooper
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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