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RE: Q on copying
- To: <xsl-list at lists dot mulberrytech dot com>
- Subject: RE: [xsl] Q on copying
- From: "Michael Kay" <mhkay at iclway dot co dot uk>
- Date: Fri, 3 Aug 2001 09:37:50 +0100
- Reply-To: xsl-list at lists dot mulberrytech dot com
After
<xsl:template name="Root">
<Root>
add
<xsl:copy-of select="@*"/>
Mike Kay
Software AG
> -----Original Message-----
> From: owner-xsl-list@lists.mulberrytech.com
> [mailto:owner-xsl-list@lists.mulberrytech.com]On Behalf Of Rosa I-Ting
> Cheng
> Sent: 03 August 2001 05:53
> To: 'xsl-list@lists.mulberrytech.com'
> 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
>
>
>
>
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>
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