[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] Generating an Xpath expression dynamically
> 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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