Re: [xsl] select distinct

["G. Ken Holman" <gkholman@xxxxxxxxxxxxxxxxxxxx>]

At 00/12/15 16:45 -0800, Richard Lander wrote:
> I will mention that the elements are not all siblings. Some are, some are
> further down. The input looks like:
>
> x
> x
> x
> y
>   x
>   x
>   x
> y
>   x
>   x
>   x
> x
> x
>
> I'm trying to get all the x elements with distinct @type values.

You can use the Meunchen method as follows to find all like type attributes 
in the document scope:

  <xsl:key name="typeatts" match="x" use="@type"/>
  ...
     <xsl:for-each select="//x[generate-id(.)=
                               generate-id(key(typeatts,@type))]">
        <A href="javascript:element('{@type}')">
          <xsl:value-of select="@type"/>
        </A>
     </xsl:for-each>

If you need to subset this within a subtree then you can incorporate into 
the key value the generated id of the apex of the subtree:

  <xsl:key name="typeatts" match="x"
           use="concat( generate-id(ancestor::y),' ',@type )"/>

and change the predicate accordingly.

I hope this helps (the fragment above is untested so I may have mistyped).

............. Ken

--
G. Ken Holman                    mailto:gkholman@xxxxxxxxxxxxxxxxxxxx
Crane Softwrights Ltd.             http://www.CraneSoftwrights.com/s/
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