Subject: Re: [xsl] Generating an Xpath expression dynamically From: Jeff Kenton <jkenton@xxxxxxxxxxxxx> Date: Tue, 21 Oct 2003 12:32:58 -0400 |
Hi, I am trying to use XSLT to transform a set of XML files, each type with its own set of nodes, into a legacy file format. It is best if the XSLT was generic. I've tried to reference a second XML file that contains the transformation rules. This second XML file contains the relative XPath expression for each element in the source XML file and the target location in the legacy file format. I get the relative XPath string from the second XML file into a parameter, but when I try to use that parameter in a XSLT value-of element, it outputs the relative XPath string rather than the contents of that node in the source XML file.
Does anyone know how to output the contents of an element node in a source XML file using a XPath string read in from another XML file? Thanks.
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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