Re: xsl variable - issue

I guess, the XSL Template fragment posted by you is
right.. But some error ;-might be there-; in the XSL
context calling this template (i.e. -- <xsl:template
match="book">)

I wrote the following XSL (its complete with all
respects), to see what could be wrong with *the XSL
fragment* posted by you.. the following XSL written by
me works fine, and correctly executes the <xsl:if
statement ..

<xsl:stylesheet
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
version="1.0">
<xsl:output indent="yes" encoding="iso-8859-1"/>
	
   <xsl:template match="/root">
      <xsl:apply-templates select="book"/>
   </xsl:template>

   <xsl:template match="book">
      <xsl:variable name="contents-var"
select="contents"/>
      <xsl:if test="$contents-var/@isvisible =
'true'">
        <!-- do some processing -->
      </xsl:if>
   </xsl:template>       

</xsl:stylesheet>

;-Hope you have already solve the problem;- . My
answer explains the problem, how I have understood
it..

All the best!

Regards,
Mukul


--- Mark Ivs <markivs2003@xxxxxxxxx> wrote:
> Hello,
>    Please take a look at the XML and XSL files. I
> have
> variable called contentsVar which contains a node
> set.
> For some reason, the 'if' statement is failing
> though
> the XML says isvisible = true. Not sure what I am
> doing wrong. 
> Any help will be appreciated.
> 
> Thanks.
> 
> XML:
> 
> <root>
>    <book>
>       <contents isvisible="true">
>           <chap>
> 	  <chap>
>        </contents>
>    </book>
> </root>
> 
> XSL:
> 
> <xsl:template match="book">
> <xsl:variable name="contentsVar" select="contents"/>
> .
> .
> .
> <xsl:if test="$contentsVar/@isvisible = 'true'">
> do something...
> </xsl:if>
> </xsl:template>
> 
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> 


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