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="$contents-var/@isvisible =
'true'">
<!-- do some processing -->
</xsl:if>
</xsl:template>
</xsl:stylesheet>
;-Hope you have already solve the problem;- . My
answer explains the problem, how I have understood
it..
All the best!
Regards,
Mukul
--- Mark Ivs <markivs2003(_at_)yahoo(_dot_)com> wrote:
Hello,
Please take a look at the XML and XSL files. I
have
variable called contentsVar which contains a node
set.
For some reason, the 'if' statement is failing
though
the XML says isvisible = true. Not sure what I am
doing wrong.
Any help will be appreciated.
Thanks.
XML:
<root>
<book>
<contents isvisible="true">
<chap>
<chap>
</contents>
</book>
</root>
XSL:
<xsl:template match="book">
<xsl:variable name="contentsVar" select="contents"/>
.
.
.
<xsl:if test="$contentsVar/@isvisible = 'true'">
do something...
</xsl:if>
</xsl:template>
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__________________________________
Do you Yahoo!?
The New Yahoo! Shopping - with improved product search
http://shopping.yahoo.com
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list