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RE: [xsl] Q on copying

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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
> 

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