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RE: [xsl] Generating an Xpath expression dynamically


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.
> > 
> 
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> 


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