RE: [xsl] Problem with generating references.

[<Jarno.Elovirta@xxxxxxxxx>]

Hi,

> I want to write an XSL file which results in XML file. The
> transformation
> should be like this..
> Note : For the nodes in the above xml file, no attribute is there to
> identify uniquely (No ids are there. But in the result tree,
> I want unique
> id for each element so that I can identify each element uniquely).

Use generate-id(), unless you want the ID to mean something and not just be
unique within that single transformation process.

> 1. Divide the elements into groups.
> 2. While dividing, create the references to the elements in
> the current
> group which are going to be placed in another group.
>
> I want the output XML like this...
> <A-Group>
>          <A1>
>                   <reference id="(id of B1)" />

Say the id of B1 is X

>                   <reference id="(id of B2)" />
>          </A1>
> </A-Group>
> <B-Group>
>          <B1/>

How do you know this is X without marking it up?

	<xsl:key name="group" match="*" use="substring(name(), 1, 1)"/>

	<xsl:template match="/">
		<xsl:for-each select="descendant::*[generate-id() = generate-id(key('group',
substring(name(), 1, 1)))]">
			<xsl:variable name="group" select="substring(name(), 1, 1)"/>
			<xsl:element name="{$group}-Group">
				<xsl:for-each select="key('group', $group)">
					<xsl:copy>
						<xsl:attribute name="id">
							<xsl:value-of select="generate-id()"/>
						</xsl:attribute>
						<xsl:for-each select="*">
							<reference id="{generate-id()}"/>
						</xsl:for-each>
					</xsl:copy>
				</xsl:for-each>
			</xsl:element>
		</xsl:for-each>
	</xsl:template>

Cheers,

Jarno

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