Perhaps you aren't supplying a value for the two stylesheet parameters?

Michael Kay
http://www.saxonica.com/ 

> -----Original Message-----
> From: Ann Marie Rubin [mailto:Annmarie.Rubin@xxxxxxx] 
> Sent: 05 November 2004 14:37
> To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> Subject:  XSLT - update attribute with new value
> 
> Hello David,
> 
> Thanks very much for your help.  At least I now know that 
> this is possible.
> 
> My stylesheet outputs the xml file unchanged. It does not replace the
> value with the value of $value.  I am probably doing something wrong
> with the variable declaration but don't see the problem.  The 
> stylesheet
> looks like this:
> 
> Do you see what might be wrong?
> 
> Thanks very much,
> 
> Ann Marie
> 
> <?xml version="1.0" encoding="UTF-8"?>
> <xsl:stylesheet version="1.0"
> xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
>     <xsl:output indent="yes" method="xml"/>
>     <xsl:variable name="attr" select="deployment-version"/>
>         <xsl:variable name="value" select='TEST'/>
> <xsl:param name="attr"/>
> <xsl:param name="value"/>
> 
> 
> <xsl:template match="node()|@*">
> 
> <xsl:copy>
> <xsl:apply-templates select="@*|node()"/>
> </xsl:copy>
> </xsl:template>
>  
> <xsl:template match="@*" priority="10">
> <xsl:attribute name="{name()}">
>  <xsl:choose>
>    <xsl:when test="name()=$attr"><xsl:value-of 
> select="$value"/></xsl:when>
>    <xsl:otherwise><xsl:value-of select="."/></xsl:otherwise>
>   </xsl:choose>
> </xsl:attribute>
> </xsl:template>
> 
>  
> </xsl:stylesheet>