Re: Question using xsl:if and xsl:param
---- "Williams, Chris D." <WILLIC10@xxxxxxxxxxxxxxxxxx> 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@xxxxxxxxxx - email XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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