Satish wrote:
I am working on a stylesheet to process my XML data but can't seem to
resolve duplicates successfully.
With XSLT 2.0 it sounds as if you could for-each-group, with XSLT 1.0 I
would use Muenchian grouping to eliminate duplicates.
Here's a simplified analogy of my XML
<library>
<book id="Bk1" authorId="A1" name="Book1"/>
<book id="Bk2" authorId="A2" name="Book2"/>
<author id="A2" name="Auth2" />
<book id="Bk2" authorId="A2" name="Book2"/>
<author id="A1" name="Auth1"/>
<book id="Bk4" authorId="A1" name="Book4"/>
...
</library>
Then with XSLT 1.0 I would define keys
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
<xsl:output method="html" indent="yes"/>
<xsl:key name="k1" match="book" use="@authorId"/>
<xsl:key name="k2" match="book" use="concat(@authorId, '|', @id)"/>
<xsl:variable name="authorsList">
<xsl:apply-templates select="//author" mode="group"/>
</xsl:variable>
<xsl:template match="/">
<html>
<xsl:copy-of select="$authorsList"/>
<body>
<table>
<tr>
<td>Library Name</td>
...
</tr>
</table>
</body>
</html>
</xsl:template>
<xsl:template match="author" mode="group">
<LibraryAuthor authName="{@name}">
<xsl:apply-templates
select="key('k1', @id)[generate-id() =
generate-id(key('k2',
concat(current()/@id, '|', @id))[1])]" mode="group"/>
</LibraryAuthor>
</xsl:template>
<xsl:template match="book" mode="group">
<LibraryBook name="{@name}" authId="{@authorId}"/>
</xsl:template>
</xsl:stylesheet>
The <xsl:copy-of select="$authorsList"/> is just for debugging, you
probably want to process that variable further although with XSLT 1.0
you will then need to use exsl:node-set or similar first.
--
Martin Honnen --- MVP Data Platform Development
http://msmvps.com/blogs/martin_honnen/
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