--- Artur Matysiak wrote:
hi
does anybody know how to convert the following XML:
<elems>
<elem id=1>
<elem id=1/>
</elem>
<elem id=1>
<elem id=2/>
</elem>
<elem id=2>
<elem id=1/>
</elem>
<elem id=2>
<elem id=2/>
</elem>
</elems>
to the form
<elems>
<elem id=1>
<elem id=1/>
<elem id=2/>
</elem>
<elem id=2>
<elem id=1/>
<elem id=2/>
</elem>
</elems>
thanks Artur
Hi Artur,
Here's a solution using the Muenchian method for grouping. Note that it
is used in a double-nested manner.
source.xml (corrected your original non-well-formed xml):
----------
<elems>
<elem id="1">
<elem id="1"/>
<elem id="2"/> </elem>
<elem id="1">
<elem id="2"/> </elem>
<elem id="1">
<elem id="3"/> </elem>
<elem id="2">
<elem id="1"/> </elem>
<elem id="2">
<elem id="2"/> </elem>
</elems>
stylesheet:
----------
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
<xsl:output omit-xml-declaration="yes" indent="yes"/>
<xsl:key name="kElem" match="elem[not(parent::elem)]" use="@id"/>
<xsl:key name="kElem2" match="elem/elem"
use="concat(../@id, ':', @id)"/>
<xsl:template match="/">
<elems>
<xsl:for-each select="//elem[generate-id()
=
generate-id(key('kElem',
@id)[1]
)
]">
<xsl:copy>
<xsl:copy-of select="@*"/>
<xsl:for-each
select="//elem/elem[generate-id()
=
generate-id(key('kElem2',
concat(current()/@id,
':',
@id
)
)[1]
)
]">
<xsl:copy-of select="."/>
</xsl:for-each>
</xsl:copy>
</xsl:for-each>
</elems>
</xsl:template>
</xsl:stylesheet>
Result:
------
<elems>
<elem id="1">
<elem id="1"/>
<elem id="2"/>
<elem id="3"/>
</elem>
<elem id="2">
<elem id="1"/>
<elem id="2"/>
</elem>
</elems>
Hope this helped.
=====
Cheers,
Dimitre Novatchev.
http://fxsl.sourceforge.net/ -- the home of FXSL
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