Subject: RE: [xsl] Generating an Xpath expression dynamically From: "Michael Kay" <mhk@xxxxxxxxx> Date: Tue, 21 Oct 2003 17:43:31 +0100 |
You're looking for the xx:evaluate() function which is defined in EXSLT and is available in a number of popular processors. Though the specs are likely to vary slightly in terms of the context for evaluating the expression, e.g. whether it allows variables or not. Michael Kay > -----Original Message----- > From: owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx > [mailto:owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx] On Behalf Of > Anderson, Rick A > Sent: 21 October 2003 17:03 > To: 'XSL-List@xxxxxxxxxxxxxxxxxxxxxx' > Subject: [xsl] 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 > XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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