Subject: [xsl] order problem From: "Shadab" <mohammad.shadab@xxxxxxxxxxx> Date: Fri, 27 Jun 2003 13:32:41 +0530 |
> Hi, > > I have this element hierarchy > > Source:: > root > body > record1(*) > record2(*) > > Destination:: > root1 > body1 > record3(*) > record4(*) > > 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
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