Subject: Re: [xsl] Q on copying From: "Alexander Gutman" <gutman@xxxxxxxxxxxxxxx> Date: Fri, 3 Aug 2001 12:13:37 +0700 |
Hello. Rosa I-Ting Cheng wrote: > 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"/> Replace all "Info"s with "info"s. > <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> Insert the following here: <xsl:copy-of select="@*"/> > <xsl:for-each select="info"> > <xsl:sort select="sort"/> Replace select="sort" with select="@sort" and add data-type="number" if you treat the sort's values as numbers rather than strings. > <xsl:copy-of select="."/> > </xsl:for-each> > <xsl:for-each select="other"> > <xsl:sort select="sort"/> Do the same here. > <xsl:copy-of select="."/> > </xsl:for-each> > </Root> > </xsl:template> -- Alexander E. Gutman XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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