Subject: Re: [xsl] problem defining param value From: "G. Ken Holman" <gkholman@xxxxxxxxxxxxxxxxxxxx> Date: Tue, 08 Jan 2013 15:53:38 -0500 |
used the following syntax in the NEW_STYLESHEET.xsl:
<xsl:variable name="fork1"> <xsl:value-of select="string that returns either 'FORK' or 'SPOON'"/> </xsl:variable>
<xsl:param name="OUTSIDE_PARAM"> <xsl:if test="$fork1 = 'FORK'"> <xsl:value-of select="string that returns 'VALUE1'"/> </xsl:if> <xsl:if test="$fork1 = 'SPOON'"> <xsl:value-of select="string that returns 'VALUE2'"/> </xsl:if> </xsl:param>
Unfortunately, the stylesheet is no longer recognizing that a param has been defined at all, and is always defaulting to VALUE3, as determined by the "otherwise" section. It seems to think that $OUTSIDE_PARAM is always empty, even though it's not. Am I missing something completely obvious?
<xsl:message>Fork: <xsl:value-of select="$fork1"/></xsl:message> <xsl:message>Param: <xsl:value-of select="$OUTSIDE_PARAM"/></xsl:message>
<xsl:comment>Fork: <xsl:value-of select="$fork1"/> Param: <xsl:value-of select="$OUTSIDE_PARAM"/></xsl:comment>
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