RE: [xsl] dinamic sorting question

["Aniceto Lopez" <aniceto@xxxxxxxxxxxxx>]

Thanks Michael, this @*[...] is working good

Ani



-----Mensaje original-----
De: owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx
[mailto:owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx]En nombre de Michael Kay
Enviado el: lunes, 16 de diciembre de 2002 11:04
Para: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Asunto: RE: [xsl] dinamic sorting question


> ok Michael here are the samples
> 
> -----XML
> <sites-list orderedcol="cod" order="descending" type="number">
>   <site id="4" cod="100" folder-name="SI01"/>
>   <site id="2" cod="0.5" folder-name="SI35"/>
>   <site id="5" cod="50" folder-name="SI05"/>
>   <site id="1" cod="25" folder-name="SI44"/>
>   <site id="3" cod="0.25" folder-name="SI05"/>
> </sites-list>
> 
> 
> -----XSL
> <xsl:variable name="queelemento" select="sites-list/@orderedcol"/>
> <xsl:variable name="queorden" select="sites-list/@order"/>
> <xsl:variable name="tipoelemento" select="sites-list/@type"/>
> 
> <xsl:for-each select="sites-list/site">
>  <xsl:sort select="*[name()=$queelemento]" order="{$queorden}"

You are selecting child elements rather than attributes. Despite its
name, $queelemento actually names an attribute. Use select="@*[...] in
place of select="*[...]".

Michael Kay
Software AG
home: Michael.H.Kay@xxxxxxxxxxxx
work: Michael.Kay@xxxxxxxxxxxxxx 


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