Thanks for the link, however it stills seems a little over my head. How to
get all the unique <ref/> elements?
<!-- XML snippet -->
<document>
<section>
<para>text text text text <ref name="a1" /> text text text.</para>
</section>
<section>
<para>text text text text <ref name="y1" /> text text text.</para>
</section>
<section>
<para>text text text text <ref name="b4" /> text text text.</para>
</section>
<section>
<list type="number">
<listitem><ref name="a1" /></listitem>
<listitem><ref name="ko" /></listitem>
<listitem><ref name="7u" /></listitem>
<listitem><ref name="y1" /></listitem>
<listitem><ref name="b4" /></listitem>
</list>
</section>
</document>
<!-- Desired output -->
<ref name="a1" />
<ref name="y1" />
<ref name="b4" />
<ref name="ko" />
<ref name="7u" />
I assume my xsl:key element would look like
<xsl:key name="name-of-unique-list" match="ref" use="@name" />
but then how do I make sure I am matching on all ref elements regardless of
where they are located in the document?
what would my xsl:template @match value be?
what would my xsl:for-each @select value be?
<xsl:key name="name-of-unique-list" match="ref" use="@name" />
<xsl:template match="???">
<xsl:for-each select="???">
???
</xsl:for-each>
</xsl:template>
Thanks for you help
Noel
----- Original Message -----
From: "Dimitre Novatchev" <dnovatchev(_at_)yahoo(_dot_)com>
To: <xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com>
Sent: Monday, September 16, 2002 1:28 PM
Subject: [xsl] Re: select distinct
[skip]
I would like to get all the unique <ref /> elements.
<ref name="a1" />
<ref name="a2" />
I am not sure what my xsl should look like to accomplish this.
Thanks
Noel
Read about the Muenchian method for grouping. One place where you can
find this explained is:
http://www.topxml.com/code/default.asp?p=3&id=v20010129150851
=====
Cheers,
Dimitre Novatchev.
http://fxsl.sourceforge.net/ -- the home of FXSL
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