Use group-by="(@associationName, '')[1]"
so that elements with no @associationName get a grouping key of "".
Michael Kay
http://www.saxonica.com/
-----Original Message-----
From: Annina(_dot_)Hirschi-Wyss(_at_)swisstopo(_dot_)ch
[mailto:Annina(_dot_)Hirschi-Wyss(_at_)swisstopo(_dot_)ch]
Sent: 14 June 2007 09:59
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [xsl] grouping-problem
Hi,
I'm working with
XSLT 2
Saxon 8
And I am a newbie with XSLT...
----
Here is the XML file
"stats.xml"
<statistic>
<class name="CI_Citation" associationName="citation"
count="50"> </class> <class name="CI_Citation"
associationName="title" count="100"> </class> <class
name="MD_Metadata" count="300"> </class> <class
name="MD_Metadata" associationName="parentId" count="3">
</class> <class name="CI_Citation" associationName="citation"
count="60"> </class> <class name="CI_Citation"
associationName="title" count="200"> </class> <class
name="MD_Metadata" count="50"> </class> <class
name="MD_Metadata" associationName="parentId" count="10">
</class> [...] </statistic>
----
Output should be a csv-textfile like:
CI_Citation citation 110
title 300
MD_Metadata 350
parentId 13
----
I managed quite well, but only did my XSLT not consider the
class MD_Metadata without the "associationName":
"stats.xsl"
<?xml version="1.0" encoding="ISO-8859-1"?> <xsl:stylesheet
version="2.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output encoding="ISO-8859-1"/>
<xsl:output method="text"/>
<xsl:template match="/">
<xsl:apply-templates select="statistic"/>
</xsl:template> [...]
<xsl:for-each-group select="class" group-by="@name">
<xsl:value-of select="current-grouping-key()"/>
<xsl:for-each-group select="current-group()"
group-by="@associationName" >
<xsl:text>;</xsl:text>
<xsl:value-of select="current-grouping-key()"/>
<xsl:text>;</xsl:text>
<xsl:value-of
select="sum(current-group()/@count)"/>
</xsl:for-each-group>
</xsl:for-each-group>
[...]
----
Is there a possibility to list the class MD_Metadata even if
there is no "associationName"? Thanks for your reply!
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