Re: [xsl] [xsl 1.0] howto merge branches by name

["G. Ken Holman" <gkholman@xxxxxxxxxxxxxxxxxxxx>]

At 2011-12-14 22:20 +0200, Adrian Herscu wrote:
> 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,

You don't say what you want help with, or where you are having 
problems.  Nor do you say what limitations you have on which version 
of XSLT to use.  That makes it difficult to advise you on how to help 
with your understanding.

Below is an XSLT 2 solution that I hope helps.

. . . . . . . . . . Ken

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:output indent="yes"/>

<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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