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RE: [xsl] Finding distinct nodes using a function in XSLT 2.0

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 Dimitre,

That works nicely.

However, is there any difference between
<xsl:sequence select="."/>
(which works) rather than say
<xsl:sequence select="current-group()[1]"/>
which also works, in the overall template logic?

thanks,
Scott

-----Original Message-----
From: Dimitre Novatchev [mailto:dnovatchev@xxxxxxxxx]
Sent: Tuesday, April 17, 2007 4:51 PM
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Subject: Re: [xsl] Finding distinct nodes using a function in XSLT 2.0

THe following transformaton s a straightforward solutionto this proble:

<xsl:stylesheet version="2.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
>

 <xsl:output omit-xml-declaration="yes" indent="yes"/>
  <xsl:variable name="vDoc1" as="document-node()"
    select="document('file:///C:/XSLT/Tests/docFoo1.xml')"/>

  <xsl:variable name="vDoc2" as="document-node()"
    select="document('file:///C:/XSLT/Tests/docFoo2.xml')"/>

  <xsl:template match="/">
   <foo>
    <xsl:for-each-group select="($vDoc1 | $vDoc2)/*/bar"
      group-by="@id">
      <xsl:sort select="current-grouping-key()"/>
      <xsl:sequence select="."/>
    </xsl:for-each-group>
   </foo>
  </xsl:template>
</xsl:stylesheet>




--
Cheers,
Dimitre Novatchev
---------------------------------------
Truly great madness cannot be achieved without significant intelligence.
---------------------------------------
To invent, you need a good imagination and a pile of junk
-------------------------------------
You've achieved success in your field when you don't know whether what
you're doing is work or play


On 4/17/07, Scott Lynch <slynch@xxxxxxxxxx> wrote:
>  This probably isn't the easiest way to this (suggestions welcome),
but
> I'm collecting data from two similarly structured source documents and
> wanting to build a set of distinct nodes based on a certain attribute
> value. Example XML:
>
> doc1.xml
> <foo>
>  <bar id="a" />
>  <bar id="b" />
>  <bar id="c" />
>  <bar id="d" />
>  <bar id="g" />
> </foo>
>
> doc2.xml
> <foo>
>  <bar id="a" />
>  <bar id="b" />
>  <bar id="c" />
>  <bar id="d" />
>  <bar id="e" />
>  <bar id="f" />
> </foo>
>
> Desired resulting node set:
>
> <foo>
>  <bar id="a" />
>  <bar id="b" />
>  <bar id="c" />
>  <bar id="d" />
>  <bar id="e" />
>  <bar id="f" />
>  <bar id="g" />
> </foo>
>
> I'm required to provide a template function (or equivalent set of
> templates which returns a node set for use by various for-each
> constructs in other processors) which accepts the xpath (e.g.
/foo/bar)
> and an index attribute (e.g. "id") as parameters and returns a node
set
> as seen above.
>
> I've managed to write a function which generates a sorted union (I
think
> I need them sorted to find the distinct values, correct? sorry, newbie
> here..), but have yet to find a way to trim the results set so that it
> contains only distinct members. Here's what I have so far:
>
>  <xsl:variable name="compare_file" select="document('doc2.xml')"/>
>  <xsl:variable name="current_file" select="document('doc1.xml')"/>
>
>  <xsl:function name="con:distinct_rows">
>    <!-- Selects the distinct rows from the source docs and -->
>    <!-- comparison doc                                     -->
>    <xsl:param name="path"/>
>    <xsl:param name="key"/>
>    <xsl:variable name="source_rows"
> select="$current_file/saxon:evaluate($path)"/>
>    <xsl:variable name="compare_rows"
> select="$compare_file/saxon:evaluate($path)"/>
>    <xsl:variable name="sorted_union_rows">
>      <xsl:for-each select="$source_rows | $compare_rows">
>        <xsl:sort data-type="text" select="./saxon:evaluate($key)"/>
>        <xsl:sequence select="."/>
>      </xsl:for-each>
>    </xsl:variable>
>    <xsl:sequence select="$sorted_union_rows"/>
>    <!-- ??? distinct node selector instead ??? -->
>  </xsl:function>
>
>
> Once I have the sorted union, I've tried various following/previous
> sibling selectors but have yet to find a command which generates the
> distinct node set (I can find the unique nodes based on the "id" but
> that's not really helping me).
>
> Any suggestions?
>
> thanks a lot!
> Scott Lynch

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