Subject: RE: [xsl] using variable in xsl:include ? From: "Michael Kay" <mhk@xxxxxxxxx> Date: Thu, 26 Feb 2004 15:12:34 -0000 |
It doesn't work because expanding xsl:include is the first thing the XSLT processor does, long before it ever starts reading a source document. Like other programming languages, XSLT stylesheets are compiled first, then executed, and you can't change the program in the middle of the execution phase. At another level, it doesn't work because the content of the href attribute is a URI, not an XPath expression. Michael Kay # -----Original Message----- # From: owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx [mailto:owner-xsl- # list@xxxxxxxxxxxxxxxxxxxxxx] On Behalf Of Nikolas Nehmer # Sent: 26 February 2004 11:11 # To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx # Subject: [xsl] using variable in xsl:include ? # # Hi, # # is it possible to use a variable in xsl:include like e.g. # <xsl:variable name="home_directory" # select="document('../../XML/config.xml')//iese.config:home_directory"/> # <xsl:include # href="concat('file://',$home_directory,'class/Person/XSL/person_short.xs # l')"/> # # This example does not work! Does anyone know why? The parser says # something like wrong syntax for pathnames,... Any suggestions? # # Best regards, # Nick # # # XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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