Subject: Re: [xsl] Contains(@href. '/') returns false From: Jeni Tennison <jeni@xxxxxxxxxxxxxxxx> Date: Fri, 8 Mar 2002 12:39:32 +0000 |
Hi Nilesh, You've forgotten some quotes in your call: > <xsl:call-template name="lastIndexOf"> > <xsl:with-param name="string" select="substring-before(@href, '#')" /> > <xsl:with-param name="char" select="/" /> > </xsl:call-template> should be: <xsl:call-template name="lastIndexOf"> <xsl:with-param name="string" select="substring-before(@href, '#')" /> <xsl:with-param name="char" select="'/'" /> </xsl:call-template> otherwise you're passing the value of the root node as the $char parameter. Cheers, Jeni --- Jeni Tennison http://www.jenitennison.com/ XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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