xslt1.0 (firefox): counting nodes with different value, muenchian-grouping?

Question

Each new problem that I think I will be able to solve, it turns out that I can't. So, with the following XML I would like to know how many different nodes are there that have a different @num attribute. Perhaps easier to understand looking at the wished output.

Initial XML code

<data>
    <prot seq="AAA">
        <node num="2">1345</node>
        <node num="2">11245</node>
        <node num="2">112w45</node>
        <node num="7">88885</node>
    </prot>
    <prot seq="BBB">
        <node num="2">678</node>
        <node num="2">456</node>
        <node num="7">4w56</node>
        <node num="7">6666</node>
    </prot>
    <prot seq="CCC">
        <node num="2">111</node>
        <node num="2">222</node>
        <node num="2">22w2</node>
        <node num="7">333</node>
        <node num="10">3433</node>
    </prot>
</data>

And the wished output, so that it expresses how many different "num"s are there

<root>
<num>2</num>
<num>7</num>
<num>10</num>
</root>

I guess it can be done (as it seems to be always the case) with muenchian grouping. I just can't see it.

Thanks!

Answer 1

I'm new to Meunchian, too. Here's my solution:

<xsl:stylesheet version="1.0"
                xmlns:xsl="http://www.w3.org/1999/XSL/Transform">

  <xsl:output method="xml" indent="yes"/>
  <xsl:key name="dupes" match="node" use="@num"/>
  <xsl:template match="/">
    <root>
      <xsl:apply-templates select="//node[generate-id() =
           generate-id(key('dupes', @num)[1])]"/>
    </root>
  </xsl:template>

  <xsl:template match="node">
    <num val="{.}">
      <xsl:value-of select="@num"/>
    </num>
  </xsl:template>
</xsl:stylesheet>

When I I run this with xsltproc:

~ zacharyyoung$ xsltproc so.xsl so.xml
<?xml version="1.0"?>
<root>
  <num val="1345">2</num>
  <num val="88885">7</num>
  <num val="3433">10</num>
</root>

I added the val="{.}" bit to show which node is being used from the key grouping. If we change ...key('dupes', @num)[1]... to ...key('dupes', @num)[last()]... we can see the difference here:

<root>
  <num val="22w2">2</num>
  <num val="333">7</num>
  <num val="3433">10</num>
</root>

For every group of nodes with corresponding values (2,7,10, etc...), the last() node in each group is selected, versus the first [1] in the previous example.

I hope this helps.