Re: [xsl] Data sequence

[Mukul Gandhi <gandhi.mukul@xxxxxxxxx>]

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@xxxxxxxxxxxxxx> 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