Subject: RE: Re: [xsl] different first element in a list From: cknell@xxxxxxxxxx Date: Mon, 24 Feb 2003 13:12:17 -0500 |
> [Lorenzo De Tomasi ] > how can I obtain an ordered list like this > ... > Rendering > > list: 1 > 2 > 3 > 4 > 5 Your example document has the values in sorted order already, so there is no need to sort them again in the XSLT. This stylesheet will do what you ask: <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:output method="html" indent="yes" encoding="UTF-8" /> <xsl:strip-space elements="*" /> <xsl:template match="/"> <table> <xsl:apply-templates select="xml/element" /> </table> </xsl:template> <xsl:template match="xml/element"> <xsl:choose> <xsl:when test="position() = 1"> <tr><td>list:</td><td><xsl:value-of select="." /></td></tr> </xsl:when> <xsl:otherwise> <tr><td> </td><td><xsl:value-of select="." /></td></tr> </xsl:otherwise> </xsl:choose> </xsl:template> </xsl:stylesheet> If the document does not contain the elements in sorted order, you could change this: <xsl:apply-templates select="xml/element" /> to this: <xsl:apply-templates select="xml/element"> <xsl:sort select="." data-type="number" /> </xsl:apply-templates> -- Charles Knell cknell@xxxxxxxxxx - email XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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