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Re: [xsl] xsl variable - issue

From: Mukul Gandhi <mukulgw3@--------->
Date: 10/25/2003 7:18:00 AM
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

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:output indent="yes" encoding="iso-8859-1"/>
   <xsl:template match="/root">
      <xsl:apply-templates select="book"/>

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


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

All the best!


--- 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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