Re: [xsl] Question using xsl:if and xsl:param

["Charles Knell" <cknell at onebox dot com>]

---- "Williams, Chris D." <WILLIC10@mail.northgrum.com> wrote:
> I am trying to do a xsl:if with a value that was assigned using xsl:param.
> How do I reference it...here is what I have tried some far using Xalan.
> 
> <xsl:stylesheet version="1.0"
> xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
> xmlns:xs="http://www.w3.org/2001/XMLSchema";>
>   <xsl:param name="SHOW_IMAGES"/>
>   <xsl:template match="/">
>   <xsl:if test="$SHOW_IMAGES == 'TRUE'">


I can't say that I know the particulars of the error message you report,
but I have repeatedly made a mistake when passing a string as a parameter
to a stylesheet, so that I suspect that your problem may also be mine.

I'm sure one of the experts will correct me if I don't have this precisely
correct, but as I understand it, a string passed as a parameter which
is enclosed in only one set of quotation marks will be interpreted as
a node-set, while a doubly-quoted string will be interpreted as a literal
string.

Look at the program that passes the parameter and see if your parameter
is enclosed in two sets of quotation marks. If not, add a second set
and try the transformation again. (i.e. "TRUE" -> "'TRUE'")
-- 
Charles Knell
cknell@onebox.com - email
 

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