xsl-list
[Top] [All Lists]

RE: Generating an Xpath expression dynamically

2003-10-21 09:43:31
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(_at_)lists(_dot_)mulberrytech(_dot_)com 
[mailto:owner-xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com] On Behalf Of 
Anderson, Rick A
Sent: 21 October 2003 17:03
To: 'XSL-List(_at_)lists(_dot_)mulberrytech(_dot_)com'
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



<Prev in Thread] Current Thread [Next in Thread>