[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] Re: can the Muenchian method do this?
On 1/4/06, dan@xxxxxxxxxxxxx <dan@xxxxxxxxxxxxx> wrote: > Hi Andrew, > > Thanks for the help but unfortunately this is not giving me the results I > desire. When I run that, it returns an empty set. It also takes a long > time to run. There is no way to use the Muenchian method but base it on > another value in the node? You didn't say you had a large input set or that performance was important... This stylesheet uses two keys, one to filter based on genre, the other to group by Artist (which is the Muenchian method :) <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:key name="song-by-genre" match="song" use="Genre"/> <xsl:key name="song-by-artist" match="song" use="Artist"/> <xsl:param name="genre" select="'Rap'"/> <xsl:template match="/"> <xsl:apply-templates select="key('song-by-genre', $genre)"/> </xsl:template> <xsl:template match="song"> <xsl:if test="generate-id() = generate-id(key('song-by-artist', Artist)[1])"> <xsl:value-of select="Artist"/> <xsl:if test="position() != last()">, </xsl:if> </xsl:if> </xsl:template> </xsl:stylesheet> cheers andrew
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