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

["NILESH PATEL" <jayganesh786@xxxxxxxxxxx>]

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@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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