Source XML:

<books>
<author>
<fname>William</fname>
<lname>Shakespeare</lame>
<bookName>Merchant of Venice</bookName>
<bookRank>One</bookRank>
</author>
<author>
<fname>William</fname>
<lname>Shakespeare</lame>
<bookName>As You Like It</bookName>
<bookRank>Two</bookRank>
</author>
</books>

Needs to be transformed to:

<books>
<author>
<fname>William</fname>
<lname>Shakespeare</lname>
<books>
<book>
<bookName>Merchant of Venice</bookName>
<bookRank>One</bookRank>
</book>
<book>
<bookName>As You Like It</bookName>
<bookRank>Two</bookRank>
</book>
</books>
</author>
</books>

Can you give me the XSL for this transformation?

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Try this stylesheet. Note it uses version2.0, so may not be compatible with some browsers.

<xsl:stylesheet version="2.0"
xmlnssl="http://www.w3.org/1999/XSL/Transform">
<xsl:output indent="yes"/>
<xsl:template match="books">
<books>
<xsl:for-each-group select="author"
group-by="concat(fname,'/',lname)">
<author>
<xsl:copy-of select="fname,lname"/>
<books>
<xsl:for-each select="current-group()">
<book>
<xsl:copy-of select="bookName,bookRank"/>
</book>
</xsl:for-each>
</books>
</author>
</xsl:for-each-group>
</books>
</xsl:template>
</xsl:stylesheet>

__________________

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try posting what you have already done! this looks like homework!

it's also the exercise in the last xsl book I read!

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