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RE: [xsl] order problem
Subject: RE: [xsl] order problem From: "Michael Kay" <mhk@xxxxxxxxx> Date: Fri, 27 Jun 2003 09:51:25 +0100 |
> > > > I have this element hierarchy > > > > Source:: > > root > > body > > record1(*) > > record2(*) > > > > Destination:: > > root1 > > body1 > > record3(*) > > record4(*) Actually, in the text of your message, you suggest that your source has a quite different structure, namely: root body ( record1 | record2 ) * This is a classic case for xsl:apply-templates. <xsl:apply-templates select="record1 | record2"/> <xsl:template match="record1"> ... </xsl:template> <xsl:template match="record2"> ... </xsl:template> > > > > Here occurence of record1 and record2 can occur any number of > > times.This will be my source xml on which the transform is to be > > applied.Following is the xsl > > > > <xsl:template match="/" name="root"> > > <xsl:element name="root1"> > > <xsl:element name="body1"> > > <xsl:for-each select="/root/body/record1"> > > <xsl:element name="record3"><xsl:value-of > > select="concat(.,'1')"/></xsl:element> > > </xsl:for-each> > > <xsl:for-each select="/root/body/record2"> > > <xsl:element name="record4"><xsl:value-of > > select="concat(.,'2')"/></xsl:element> > > </xsl:for-each> > > </xsl:element> > > </xsl:element> > > </xsl:template> > > > > Only problem would be that i need the order of records to be same in > output > > as in input and input of records can be in random order,viz > > <record1><record1><record2><record1><record2><record1>.But with the > > two for-loops this will break the sequence. > > > > How can this be done. > > > > Any help on this would be deeply appreciated. > > > > Thanks, > > Shadab > > > > > XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list > XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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