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[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] Re: Selecting "distinct" elements
This is a simple grouping problem. An efficient solution is using the
Muenchian method for grouping.
Read about it at:
http://www.topxml.com/code/default.asp?p=3&id=v20010129150851
or at:
http://jenitennison.com/xslt/grouping/index.html
=====
Cheers,
Dimitre Novatchev.
http://fxsl.sourceforge.net/ -- the home of FXSL
"Marcus Andersson" <marcan@xxxxxxx> wrote in message
news:000001c384c6$960ef6e0$0400a8c0@xxxxxxxxxxx
> Hi
> I want to select one of each element with a certain attribute value and
> loop over these elements in a for-each (or possibly apply-templates). Is
> this possible with a single select statement or do I have to make a more
> cumbersome solution? If possible, how would a statement solving the
> problem look like?
>
> Instance document:
> <root>
> <element value="a"/>
> <element value="b"/>
> <element value="a"/>
> <element value="c"/>
> <element value="b"/>
> <element value="d"/>
> <element value="b"/>
> </root>
>
> XSLT Template:
> <xsl:template match="root">
> <xsl:for-each select="[insert your favourite statement here]">
> <anotherElt value="{@value}"/>
> </xsl:for-each>
> </xsl:template>
>
> Resulting document:
> <root>
> <anotherElt value="a"/>
> <anotherElt value="b"/>
> <anotherElt value="c"/>
> <anotherElt value="d"/>
> </root>
>
> Thanks,
>
> /Marcus
>
>
> XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
>
>
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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