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Subject: Beyond XSL's capabilities? From: "Blanchette, Larry" <Larry.Blanchette@xxxxxxxxxxxxx> Date: Tue, 16 Mar 1999 13:55:20 -0600 |
I've been unable to reach a solution for the following example where I would like to group a set of elements based on an attribute value, this would be driven by a different element's attribute value. My conclusion is you can't get there with xsl and the only recourse is to rework the schema. Is this correct? <?xml version="1.0" ?> <xsl:stylesheet xmlns:xsl="http://www.w3.org/TR/WD-xsl"> <xsl:id element="beverage" attribute="bev.type"/> <xsl:template match="drink.menu"> <html><title>Drink Menu</title> <body><xsl:apply-templates select="menu"/></body> </html> </xsl:template> <xsl:template match="menu.group"> <p><xsl:value-of select="@..."/></p> <p> <!-- how to generate the groups's list ? this doesn't work id's need to be unique <xsl:apply-templates select="id(@...)"/> looked at the following but not valid: <xsl:apply-templates select= "[/drink.menu/beverage.list/beverage@xxxxxxxx = @...]"/> <xsl:for-each select="/drink.menu/beverage.list/beverage"> in the context of beverage here so i've lost menu.group@xxxxxxxx ? --> </p> </xsl:template> </xsl:stylesheet> Sample data: <?xml version="1.0"?> <drink.menu> <beverage.list> <beverage bev.type='Beer'>Sierra Nevada</beverage> <beverage bev.type='Beer'>Newcastle</beverage> <beverage bev.type='Vodka'>Screwdriver</beverage> <beverage bev.type='Vodka'>Greyhound</beverage> </beverage.list> <menu> <menu.group bev.type='Vodka'/> <menu.group bev.type='Beer' /> </menu> </drink.menu> Thanks Larry Blanchette XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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