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

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.