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Re: [xsl] Data sequence

2010-10-22 03:34:23
Following Mike's algorithm, I think this is probably you'll exactly
need (this is one of the solutions):

<xsl:for-each select="author">
     <xsl:sort select="number(@seq)"/>
     <xsl:apply-templates select="." mode="x" />
</xsl:for-each>

<xsl:template match="author" mode="x">
     <au>
         <xsl:copy-of select="@*" />
         <xsl:copy-of select="node()" />
     </au>
</xsl:template>

This is not tested.

On Fri, Oct 22, 2010 at 1:24 PM, JS rawat <jrawat(_at_)aptaracorp(_dot_)com> 
wrote:

Hi Team,
Is there any quick idea about how to sequence the authors in the output
file.

Input
<article>
<author seq="1">author 1</author>
<author seq="3">author 3</author>
<author seq="2">author 2</author>
</article>

Desired output
<article>
<au seq="1">author 1</author>
<au seq="2">author 2</author>
<author seq="3">author 3</author>
</article>

Thanks in advance
...JSR



-- 
Regards,
Mukul Gandhi

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