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RE: Changing an attribute wherever it may occur


Sure,
<?xml version="1.0"?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
<xsl:template match="node()|@*">
	<xsl:copy>
		<xsl:apply-templates select="@* | * | comment() |
processing-instruction() | text()"/>
	</xsl:copy>
</xsl:template>
<xsl:template match="@Action">
<xsl:attribute name="Action">M</xsl:attribute>
</xsl:template>
</xsl:stylesheet>

Ciao Chris

XML/XSL Portal
http://www.bayes.co.uk/xml


> -----Original Message-----
> From: owner-xsl-list@lists.mulberrytech.com 
> [mailto:owner-xsl-list@lists.mulberrytech.com] On Behalf Of 
> mjyoungblut@mmm.com
> Sent: 27 September 2001 18:52
> To: xsl-list@lists.mulberrytech.com
> Subject: [xsl] Changing an attribute wherever it may occur
> 
> 
> Hi,
>      Is there a way to change an attribute in any occurrence 
> in a given element, including children?  I have been able to 
> do it recursively, with a little knowledge of where the 
> attributes might be, but is it possible to do it all at once? 
>  I want to assume that I don't know where the attribute will be at.
> 
> For example,
> <A>
>      <B Action='A'>...</B>
>      <C>
>           <D Action='B'>...</D>
>      </C>
>      <E>
>           <F>
>                <G Action='A'>...</G>
>           </F>
>      </E>
> </A>
> 
> Is it possible to copy <A> while changing all of the @Action 
> attributes to 'M'?
> 
> Thanks for your help in advance,
> Matt Youngblut
> 
> 
>  XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list
> 
> 


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