Re: xsl:key grouping problem

Thank you for the detialed explanation. It was very
helpful. One last thing. I would like to display my
results in 2 or 3 table columns. The way it is right
now, it will only display one column. Please see xsl
below. Thanks again.

  <xsl:key name="groups" match="country"
use="concat(country_group, 
@year, @num)"/>

  <xsl:template match="myroot">
    <table>
    <xsl:for-each select="country[generate-id() =
generate-id(key('groups', concat(country_group,
'2004', '2')))]">
      <tr><td>
         <xsl:value-of select="country_group"/>
      </td></tr>
    </xsl:for-each>
    </table>
  </xsl:template>
  
Thanks again.
Mark


--- David Carlisle <davidc@xxxxxxxxx> wrote:

> 
> 
> > Using xsl keys seems like a tough concept to
> > understand(atleast for me). Are there other
> novices
> 
> keys themselves are simple enough, but using them
> (as here) for grouping
> isn't at all obvious and is normally called (here at
> least) Muenchian
> grouping after Steve Muench who first thought of
> this.
> Jeni's site is the usual reference for a tutorial.
> 
> http://www.jenitennison.com/xslt/grouping/index.html
> 
> >  How does generate-id on country be equal to
> >  generate-id on the key? That's where I am
> confused.
> >  country[generate-id() = generate-id(key('groups',
> >  concat(country_group, '2004', '2'))) ??
> 
> 
> The trick here is that if you are on a country then
> concat(country_group, '2004', '2')
> is the key string you want to use and so
> 
> key('groups', concat(country_group, '2004', '2'))
> 
> will return all the nodes that we want to consider
> to be grouped with
> this country.
> 
> Now the test is trying to say "is this node the
> first node in this
> group" (if so start processing the group, if not do
> nothing as this node
> was already handled when the rest of the group was
> handled, on the first
> node)
> 
> so this node is .
> 
> and the first node in the current group is
> 
> key('groups', concat(country_group, '2004', '2'))[1]
> 
> so you want to know if
> 
> . is key('groups', concat(country_group, '2004',
> '2'))[1]
> 
> now in XPath 2 (draft) there is an "is" operator
> that tests node
> identity and the above line is actually valid Xpath2
> but in Xpath 1
> there is no operator. You can't "is". You can't use
> use = as
> 
> 
> . = key('groups', concat(country_group, '2004',
> '2'))[1]
> 
> would test if the string value of the current node
> was equal to the
> string value of the first node in the group, which
> isn't what we want.
> 
> however generate-id returns a unique string for each
> node so you can say
> 
> generate-id(.) = generate-id(key('groups',
> concat(country_group, '2004', '2'))[1])
> 
> and that will be true just if the current node is
> the first node in the
> group.
> 
> This idiom isn't at all obvious but it is a FAQ and
> when you've seen it
> most days on this list for 5 years you tend to take
> some syntactic
> shortcuts:
> 
> generate-id(.)
> 
> can be written
> 
> generate-id() because . is implied if you supply no
> argument.
> 
> generate-id(key('groups', concat(country_group,
> '2004', '2'))[1])
> 
> can be written
> 
> generate-id(key('groups', concat(country_group,
> '2004', '2')))
> 
> because as for most string generating functions in
> XSLT 1, if you supply
> a node set as an argument it will take the first
> node in teh set in
> document order and use that and silently ignore any
> other nodes, in
> other words [1] is implictly applied.
> 
> David
> 
>
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