Subject: Re: [xsl] To display the output depending upon condition From: Jörg Heinicke <joerg.heinicke@xxxxxx> Date: Sat, 20 Oct 2001 02:13:07 +0200 |
I don't know really what you want, my only presumption is that the output-tree of your transformation is missing. Your transformation is correct I think, you only need the tree. Therefore you must copy the given tags. <?xml version="1.0" ?> <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0"> <xsl:output encoding="iso-8859-1"/> <xsl:strip-space elements="*"/> <xsl:template match="Sections"> <xsl:copy> <xsl:apply-templates select="Messages" /> </xsl:copy> </xsl:template> <xsl:template match="Messages"> <xsl:copy> <xsl:apply-templates select="Message[@parentId=0]"/> </xsl:copy> </xsl:template> <xsl:template match="Message"> <xsl:copy> <xsl:apply-templates select="../Message[@parentId=current()/@dbMessageId]"/> </xsl:copy> </xsl:template> </xsl:stylesheet> If this is not what you want, you must explain the problem more explicitly. Joerg > But What I need is: > > 1. Now the grouping starts with parentId=0 and then > grouping all the child with them. > > 2. At this point I need to check whether that > particular parentId !=0 and if yes I need to > display that message as a particular left not a > part of tree. > > 3. The reason for this query was if I do a search > via my web interface I might get some data whose > parentId !=0, since this XSL starts with > parentId=0, I AM NOT GETTING THAT TREE AT ALL. > > Please help me to solve this problem. > > Many Thanks > Srini XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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