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RE: [xsl] Position of a Child
Subject: RE: [xsl] Position of a Child From: "Casadome, Francisco Javier" <Francisco.Casadome@xxxxxxxxxxxxxx> Date: Thu, 20 Sep 2001 21:27:06 +0200 |
Hi, Thanks for your answer, the input data I have is pretty simple: <myXML> <ContextNode> <Child1/> <Child3/> <Child4/> ... <Childn/> </ContextNode> <ContextNode> <Child1/> <Child4/> ... <Childn/> </ContextNode> ... </myXML> The problem with the solution you gave me is that this code is inside a template (that matches the contextNode), and this template calls itself recursively for a set of selected attributes. Basically what I'm doing is to filter the input xml to create an output based in a list of Childs (that this particular node may have or not). So going through every child for each one in the list and for each contextNode would be a lot of processing. That is why I was using the select"*[name()=$NodeName]" to identify the child without going through all of them. And problem about all this is that I have to treat specially childs that are in certain positions. Hope now is clearer :) Frank. -----Original Message----- From: Wendell Piez [mailto:wapiez@xxxxxxxxxxxxxxxx] Sent: jueves, 20 de septiembre de 2001 20:58 To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx Subject: Re: [xsl] Position of a Child Frank, What you mean by "position of that child inside the context node" is actually a bit problematic, possibly meaning any of a number of things. How to get it depends both on what you mean, and what your input data looks like. For starters, you could try (count(*[name()=$NodeName]/preceding-sibling::*) + 1) -- that is, 1 + the number of the node's preceding element siblings. If you have mixed content (and thus the siblings you want to count include text nodes), or if there's some other wrinkle (such as more than one child node being named $NodeName), come back. position() isn't very useful because it returns the position of the context node within the current node list, which isn't what you want. You could make it be what you want by doing something like: <xsl:for-each select="*"> <xsl:if test="name()=$NodeName"> <xsl:value-of select="."/> <xsl:value-of select="position()"/> </xsl:if> </xsl:for-each> But again, I'm making assumptions about what you really want -- and you can see this is more roundabout than simple counting the preceding siblings. Cheers, Wendell At 02:17 PM 9/20/01, you wrote: >I'm having problems getting the child's position() of the context node. >I reference the child by its name (stored in a variable), like this: > ><xsl:value-of select="*[name()=$NodeName]"/> > >What I would like to get is the position of that child inside the context >node. ====================================================================== Wendell Piez mailto:wapiez@xxxxxxxxxxxxxxxx Mulberry Technologies, Inc. http://www.mulberrytech.com 17 West Jefferson Street Direct Phone: 301/315-9635 Suite 207 Phone: 301/315-9631 Rockville, MD 20850 Fax: 301/315-8285 ---------------------------------------------------------------------- Mulberry Technologies: A Consultancy Specializing in SGML and XML ====================================================================== XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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