RE: [xsl] Is this about grouping ?

["Michael Kay" <mhkay@xxxxxxxxxxxx>]

>
> Let's start again.
> This is my original source ( with the correct indentation ) :
> <?xml version="1.0"?>
> <topic>
> 	<a>ItemA</a>
> 	<b>ItemB</b>
> 	<f>text1</f>
> 	<g>text2</g>
> 	<h>text3</h>
> 	<i>text4</i>
>
> 	<a>ItemA</a>
> 	<b>ItemB</b>
> 	<f>text5</f>
> 	<g>text6</g>
> 	<h>text7</h>
> 	<i>text8</i>
>
> 	<a>ItemA</a>
> 	<b>ItemB</b>
> 	<f>text9</f>
> 	<g>text10</g>
> 	<h>text11</h>
> 	<i>text12</i>
> </topic>
>
> and the result that I want to have  ( with the correct indentation ) :
> <?xml version="1.0"?>
> <topic>
> 	<a>ItemA</a>
> 	<b>ItemB</b>
> 		<f>text1</f>
> 		<g>text2</g>
> 		<h>text3</h>
> 		<i>text4</i>
> 		<f>text5</f>
> 		<g>text6</g>
> 		<h>text7</h>
> 		<i>text8</i>
> 		<f>text9</f>
> 		<g>text10</g>
> 		<h>text11</h>
> 		<i>text12</i>
> </topic>
>
> So, I need to group by <a> and <b>. Those elements are constant in the
> source document.


But what you haven't told us is, what is variable in the source document? A
transformation that only works on one source document is trivial; we need to
understand the structure to which all the possible source documents conform,
and giving one example isn't enough to explain this. The difficulty of this
problem depends very much on the extent to which the source documents will
vary.

Mike Kay
Software AG

> I will ignore the sort order because the original source is
> already sorted.
> If my XSL script is like this :
>
> <xsl:transform xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
> version="1.0">
> <xsl:output method="xml" indent="yes" omit-xml-declaration="no" />
>
> <xsl:key name="group" match="f | g | h | i" use="//a" />
>
> <xsl:template match="topic">
> <topic>
> 	<a>
> 		<xsl:value-of select="//a[1]"/>
> 	</a>
> 	<b>
> 		<xsl:value-of select="//b[1]"/>
> 	</b>
>
> 	<!--
>     <xsl:for-each select="key('group' ,a)">
> 		<xsl:choose>
> 			<xsl:when test="//f">
> 				<xsl:text/>    <xsl:value-of
> select="."/><xsl:text/>
> 			</xsl:when>
> 			<xsl:otherwise>
> 				<xsl:text/>    <xsl:value-of
> select="."/><xsl:text/>
> 			</xsl:otherwise>
> 		</xsl:choose>
>     </xsl:for-each>
> 	-->
>
>     <xsl:copy-of select="key('group' ,a)"/>
>
> </topic>
> </xsl:template>
>
> </xsl:transform>
>
> the result will be :
> <?xml version="1.0" encoding="UTF-8"?>
> <topic>
> <a>ItemA</a>
> <b>ItemB</b>
> <f>text1</f>
> <g>text2</g>
> <h>text3</h>
> <i>text4</i>
> <f>text5</f>
> <g>text6</g>
> <h>text7</h>
> <i>text8</i>
> <f>text9</f>
> <g>text10</g>
> <h>text11</h>
> <i>text12</i>
> </topic>
>
> The problem with that XSL using <xsl:copy-of> is that I can't
> control over
> the indentation
> from <f> to <i>. So I have to use the <xsl:for-each>.
> With <xsl:for-each>, I can control over the node-set returned
> by key(), i.e
> concatenate <f> and <g>.
> The above <xsl:for-each> is uncomplete or wrong completely.
>
> What I would like you to help me is how can I use <xsl:for-each> :
> 1. to have a proper indentation from <f> to <i> ?
> 2. to concatenate 2 elements ,i.e <f> and <g> into <new>, so the final
> result is :
> <?xml version="1.0" encoding="UTF-8"?>
> <topic>
>     <a>ItemA</a>
>     <b>ItemB</b>
>         <new>text1+text2</new>
>         <h>text3</h>
>         <i>text4</i>
>         <new>text5+text6</new>
>         <h>text7</h>
>         <i>text8</i>
>         <new>text9+text10</new>
>         <h>text11</h>
>         <i>text12</i>
> </topic>
>
> I would appreciate any help.
>
> Thanks
> Benoit Aumars.
>
>
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