RE: Xpath of an attribute

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(_at_)lists(_dot_)mulberrytech(_dot_)com
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][(_at_)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

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