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

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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

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Jeni Tennison
http://www.jenitennison.com/


 XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list

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