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[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] 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@xxxxxxxxxxxxxxxxxxxxxx Subject: 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
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