Subject: Re: [xsl] Contains(@href. '/') returns false From: "NILESH PATEL" <jayganesh786@xxxxxxxxxxx> Date: Fri, 08 Mar 2002 12:45:16 +0000 |
From: Jeni Tennison <jeni@xxxxxxxxxxxxxxxx> Reply-To: Jeni Tennison <jeni@xxxxxxxxxxxxxxxx> To: "NILESH PATEL" <jayganesh786@xxxxxxxxxxx> CC: XSL-List@xxxxxxxxxxxxxxxxxxxxxx Subject: Re: [xsl] Contains(@href. '/') returns false 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/
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