Subject: Re: [xsl] [xsl 1.0] howto merge branches by name From: "G. Ken Holman" <gkholman@xxxxxxxxxxxxxxxxxxxx> Date: Wed, 14 Dec 2011 15:27:48 -0500 |
Consider the following input XML file: ... I am trying to find a way to transform it into: ... The branches are always 3-levels deep. That is: suite > case > procedure.
Please help,
t:\ftemp>type adrian.xml <?xml version="1.0" encoding="UTF-8"?> <test> <suite name="A"> <case name="A"> <procedure name="A" /> </case> </suite> <suite name="A"> <case name="A"> <procedure name="B" /> </case> </suite> <suite name="A"> <case name="B"> <procedure name="A" /> </case> </suite> </test>
t:\ftemp>xslt2 adrian.xml adrian.xsl <?xml version="1.0" encoding="UTF-8"?> <test> <suite name="A"> <case name="A"> <procedure name="A"/> <procedure name="B"/> </case> <case name="B"> <procedure name="A"/> </case> </suite> </test> t:\ftemp>type adrian.xsl <?xml version="1.0" encoding="UTF-8"?> <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="2.0">
<xsl:template match="test"> <test> <xsl:for-each-group select="suite" group-by="@name"> <suite name="{@name}"> <xsl:for-each-group select="current-group()/case" group-by="@name"> <case name="{@name}"> <xsl:copy-of select="current-group()/procedure"/> </case> </xsl:for-each-group> </suite> </xsl:for-each-group> </test> </xsl:template>
</xsl:stylesheet> t:\ftemp>
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