RE: [xsl] Xpath of an attribute

["Animesh Sharma" <asharma@xxxxxxxxxxxxxxxx>]

This can be work out as follows:
<xsl:template match="//body/namespace/form/snip/csf/div/center/p[position()=1]/a[position()=1]/@href"/>

Regards,
Animesh

-----Original Message-----
From: Animesh Sharma 
Sent: Tuesday, April 27, 2004 2:18 PM
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Subject: [xsl] Xpath of an attribute


hi,

I want to remove a particular attribute of node. 
Is there way that I can write the template in one line something like:

<xsl:template match="//body/namespace/form/snip/csf/div/center/p[position()=1]/a[position()=1][@href]"/>

<xsl:template match="*|text()|@*">
<xsl:copy>
<xsl:apply-templates select="*|text()|@*"/>
</xsl:copy>
</xsl:template>

This removes the anchor element.. but I just want to remove the "href" attribute of element.

Regards,
Animesh