Re: [xsl] Contains(@href. '/') returns false

["NILESH PATEL" <jayganesh786 at hotmail dot com>]

You are great. Works now, such a silly thing to miss though. Didn't think 
about passing root node as $char.

Thanks.

Nilesh


>From: Jeni Tennison <jeni@jenitennison.com>
>Reply-To: Jeni Tennison <jeni@jenitennison.com>
>To: "NILESH PATEL" <jayganesh786@hotmail.com>
>CC: XSL-List@lists.mulberrytech.com
>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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