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RE: merge tree structure with no redundancy

Subject: RE: merge tree structure with no redundancy
From: "McNally, David" <David.McNally@xxxxxxxxxx>
Date: Fri, 2 Aug 2002 13:10:56 -0400
merge structure list
When in doubt, reach for keys and the Meunchian technique, both well
described in the faq.

Basically you need two keys - "me" which is used to figure out whether each
node is the first, in document order, with a particular @type and @name
pair.  The second, "parent", is used to process all the nodes who are
children of a particular @type and @name pair, regardless of where they are
in the file.  So basically the node template decides if the node is the
first with its type/name - if it isn't it does nothing.  If it is, it
outputs the element and attributes, and then applies templates to all nodes
that are "children" of that node, based on the "parent" key.

I think this works properly, but handle with care...

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
	<xsl:output method="xml" version="1.0" encoding="UTF-8"
indent="yes"/>

<xsl:key name="parent" match="node"
use="concat(parent::node/@type,'::',parent::node/@name)"/>
<xsl:key name="me" match="node" use="concat(@type,'::',@name)"/>

<xsl:template match="a">
<a>
	<xsl:apply-templates select="subscribers/subscriber/path/node"/>
</a>
</xsl:template>

<xsl:template match="node">
	<xsl:variable name="mykey" select="concat(@type,'::',@name)"/>
	<xsl:choose>
		<xsl:when test="generate-id(.) =
generate-id(key('me',$mykey)[1])">
			<xsl:element name="{name()}">
				<xsl:copy-of select="@*"/>
				<xsl:if test="not(node)">
					<xsl:attribute name="id">
						<xsl:value-of
select="ancestor::subscriber/id/@id"/>
					</xsl:attribute>
				</xsl:if>
				<xsl:apply-templates
select="key('parent',concat(@type,'::',@name))"/>
			</xsl:element>
		</xsl:when>
	</xsl:choose>
</xsl:template>

</xsl:stylesheet>


Note that this will not work if type/name pairs can be reused in different
type/name "parents".  I.e.:

<node type="T1" name="N1">
	<node type="T2" name="N2">
</node>
...
<node type="T1" name="N2">
	<node type="T2" name="N2"><!-- same type and name as a previous
node, but different "parent" -->
</node>

would be bad, and you'd need to combine lots of "parent" information into
the "me" key.


D:\Work\xsl>type subscriber.xml
<?xml version="1.0" encoding="ISO-8859-1"?>
<a>
 <subscribers>
   <subscriber>
     <id="1" />
       <path>
         <node type="T1" name="N1">
           <node type="T2" name="N2">
             <node type="T3" name="N3" />
           </node>
         </node>
       </path>
     </subscriber>
   <subscriber>
     <id="2" />
     <path>
       <node type="T1" name="N1">
         <node type="T2" name="N2">
           <node type="T3" name="N4" />
         </node>
       </node>
     </path>
   </subscriber>
 </subscribers>
</a>

D:\Work\xsl>saxon subscriber.xml subscriber.xslt
<?xml version="1.0" encoding="UTF-8"?>
<a>
   <node type="T1" name="N1">
      <node type="T2" name="N2">
         <node type="T3" name="N3" id="1"/>
         <node type="T3" name="N4" id="2"/>
      </node>
   </node>
</a>

Hope this helps,
David.
--
David McNally            Moody's Investors Service
Software Engineer        99 Church St, NY NY 10007 
David.McNally@xxxxxxxxxx            (212) 553-7475 


> -----Original Message-----
> From: Benoit Guyon [mailto:ben@xxxxxxxxxx]
> Sent: Friday, August 02, 2002 11:43 AM
> To: 'XSL-List@xxxxxxxxxxxxxxxxxxxxxx'
> Subject:  merge tree structure with no redundancy
> 
> 
> hello,
> 
> I have a XML file with nodes that are representing a tree 
> structure. i want
> to modify this file (with a XSL stylesheet) to get a new XML 
> file with a
> unified tree structure that don't have redundancy.
> 
> the original file :
> 
> <?xml version="1.0" encoding="ISO-8859-1"?>
> <a>
>  <subscribers>
>    <subscriber>
>      <id="1" />
>        <path>
>          <node type="T1" name="N1">
>            <node type="T2" name="N2">
>              <node type="T3" name="N3" />
>            </node>
>          </node>
>        </path>
>      </subscriber>
>    <subscriber>
>      <id="2" />
>      <path>
>        <node type="T1" name="N1">
>          <node type="T2" name="N2">
>            <node type="T3" name="N4" />
>          </node>
>        </node>
>      </path>
>    </subscriber>
>  </subscribers>
> </a>
> 
> the wanted file :
> 
> <?xml version="1.0" encoding="ISO-8859-1"?>
> <a>
>  <node type="T1" name="N1">
>    <node type="T2" name="N2">
>      <node type="T3" name="N3" id="1" />
>      <node type="T3" name="N4" id="2" />
>    </node>
>  </node>
> </a>
> 
> I don't know how to make that ... can i test output to detect 
> if a node
> (identified with type and name atributes) is
> already present ?
> 
> when i'm in the second <subscriber> node, how can i detect 
> that i already
> wrote the ancestor nodes, and how can i write the final node 
> under the right
> ancestor ?
> 
> does anybody have an idea or a general algorithm to help me ?
> 
> best regards
> ben
> 
> PS : i use MSXML, i don't have the saxon:path() :/ ;)
> 
>  XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list
> 


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