Subject: Re: [xsl] sequential numbering in xslt From: a kusa <akusa8@xxxxxxxxx> Date: Mon, 11 Jan 2010 14:28:51 -0600 |
I worked with the solution that Jim was kind enough to lay out <xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:fo="http://www.w3.org/1999/XSL/Format" xmlns:xs="http://www.w3.org/2001/XMLSchema" xmlns:saxon="http://saxon.sf.net/" extension-element-prefixes="saxon"> <xsl:param name="previous-number" select="1" /> <xsl:template match="/"> <xsl:call-template name="example"> <xsl:with-param name="documents" select="collection('file:///L:/input/?select=*.xml')" /> <xsl:with-param name="previous-number" select="$previous-number" /> </xsl:call-template> </xsl:template> <xsl:template name="example"> <xsl:param name="documents"> <xsl:param name="previous-number" required="yes" /> <xsl:if test="$documents[1]"> <xsl:variable name="current-number" as="xs:integer" select="xs:integer($previous-number+1)" /> <car seq="{$current-number}"> <xsl:copy-of select="@*|node()"/> </car> <xsl:call-template name="example"> <xsl:with-param name="documents" select="remove($documents, 1)" /> <xsl:with-param name="previous-number" select="$current-number" /> </xsl:call-template> </xsl:if> </xsl:template> </xsl:stylesheet> And my L:/input contains 2 XML files for now. The root element of both is <car> On Mon, Jan 11, 2010 at 2:21 PM, David Carlisle <davidc@xxxxxxxxx> wrote: > > >> When I say $document[1], it considers all files as $documents[1]. > > This is not possible. There is no data value in xpath that corresponds to > multiple documents except a sequence of document nodes, and as you can > not nest sequences, it is not possible to return a sequence of more than > one item as a result of $documents[1]. > > What expression did you actually use to give the impression that > $document[1] is all files. > > David
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