Subject: RE: [xsl] Q on copying From: "Michael Kay" <mhkay@xxxxxxxxxxxx> Date: Fri, 3 Aug 2001 09:37:50 +0100 |
After <xsl:template name="Root"> <Root> add <xsl:copy-of select="@*"/> Mike Kay Software AG > -----Original Message----- > From: owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx > [mailto:owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx]On Behalf Of Rosa I-Ting > Cheng > Sent: 03 August 2001 05:53 > To: 'xsl-list@xxxxxxxxxxxxxxxxxxxxxx' > Subject: [xsl] Q on copying > > > Can anyone please tell me how to copy XML into certain sorted > order while > keeping the attributes of the root element as is? > ie the following is my XML > > <Root att1="" att2=""> > <Info sort="2"/> > <Info sort="5"/> > <info sort="1"/> > <other sort="3"/> > <other sort="5"/> > <other sort="2"/> > </Root> > > I want to turn that XML into > <Root att1="" att2=""> > <info sort="1"/> > <Info sort="2"/> > <Info sort="5"/> > <other sort="3"/> > <other sort="4"/> > <other sort="5"/> > </Root> > > This is what I have so far.. > <xsl:template name="Root"> > <Root> > <xsl:for-each select="info"> > <xsl:sort select="sort"/> > <xsl:copy-of select="."/> > </xsl:for-each> > <xsl:for-each select="other"> > <xsl:sort select="sort"/> > <xsl:copy-of select="."/> > </xsl:for-each> > </Root> > </xsl:template> > > Thanx! > > Rosa > > > > > 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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