[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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