Subject: [xsl] telling an outer template which template to call... From: Felix Breuer <felix@xxxxxxxxxx> Date: 01 Sep 2003 22:13:05 +0200 |
Hello! I would like an outer template to call an inner template which is specified by a parameter. I.e. I would like to get the following example working: <xsl:template match="/"> <xsl:apply-templates select="foo"> <xsl:with-param name="para">bar</xsl:with-param> </xsl:apply-templates select="foo"> </xsl:template> <xsl:template match="foo"> <xsl:param name="para">moo</xsl:with-param> ... <xsl:call-template name="{$para}"/> ... </xsl:template> <xsl:template name="bar">...</xsl:template> <xsl:template name="moo">...</xsl:template> I hope the above makes it clear what I am aiming at. [Obviously the whole stylesheet is going to be a bit more complicated or such an involved solution wouldn't be necessary.] Now, when I run the above through xsltproc (libxslt) I get errors like xsl:call-template : template {$para} not found Obviously, the problem is that the parameter $para is not expanded before the call-template element is executed. My questions are: 1) Is $para supposed not to be expanded in time? Or is this a bug in libxslt? 2) Is there a way to get this stylesheet working? With libxslt or any other processor? Thanks in advance, Felix Breuer XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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