Re: Ordered union of sequences

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<?xml version="1.0" encoding="utf-8"?>
<foo>
  <seq><k/><o/><p/><c/><f/></seq>
  <seq><d/><e/><f/><g/></seq>
  <seq><k/><f/><z/></seq>
  <seq><a/><b/><c/><d/></seq>
</foo>
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--------------------------------------
<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet
 xmlns:xs="http://www.w3.org/2001/XMLSchema"
  xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
  xmlns:my="URN:my"
  version="2.0"  >

  <xsl:output
    method="xml"
    indent="yes"
    />

  <xsl:template match="/">
    <xsl:apply-templates />
  </xsl:template>

  <xsl:template match="foo">
    <xsl:copy>
      <xsl:for-each-group select="seq/*" group-by="name(.)" >
        <xsl:sort select="my:sortkey(.)"/>
        <xsl:element name="{current-grouping-key()}" />
      </xsl:for-each-group>
    </xsl:copy>
  </xsl:template>

</xsl:stylesheet>
--------------------------------------

--------------------------------------
<foo>
   <k/>
   <a/>
   <o/>
   <b/>
   <p/>
   <c/>
   <d/>
   <e/>
   <f/>
   <g/>
   <z/>
</foo>
--------------------------------------

There is an arbitrary number of sequences, sometimes

containing items

with the same name:

(k, o, p, c, f)
(d, e, f, g)
(k, f, z, o)
(a, b, c, d)

I want to create a master sequence which contains every item once,
preserving the original order.

Doesn't xsl:for-each-group over combined sequence solves the problem?

No, it puts groups in order of first appearance, which would put f before d.

Regards,

Michael Kay
http://www.saxonica.com/
http://twitter.com/michaelhkay

--
Gerrit Imsieke
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