Hi.

> -----Original Message-----
> From: owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx 
> [mailto:owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx] On Behalf Of 
> Martin Lampen
> Sent: Tuesday, May 20, 2003 10:22 AM
> To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> Subject: RE:  URL parameters in xsl
> 
> 
> Thanks for all your replies. None of these solutions work 
> though. the xsl file won't recognise the url - is this 
> because it's generated from an .asp file!
> 
> help!!!
> 
> martin

Others already gave you hints how to do it with xslt, here is how you
could do it in asp:

' Create a template
Set xslt = Server.CreateObject("Msxml2.XSLTemplate")

' Create the xsl document as free threaded
Set xslDoc = Server.CreateObject("Msxml2.FreeThreadedDOMDocument")

' load the stylesheet
xslDoc.async = false
xslDoc.load "stylesheet.xsl"

' set the templates stylesheet to yours
set xslt.stylesheet = xslDoc

' Create the xml document
set xmlDoc = Server.CreateObject("Msxml2.DOMDocument")

' load the xml
xmlDoc.async = false
xmlDoc.load "file.xml"

' Create the processor
Set xslProc = xslt.createProcessor

' set the processor imput to your xml doc
xslProc.input = xmlDoc

' set the output, probably to the Response object
Xslproc.output = Response

' add the parameters
xslProc.addParameter "uid", Request.QueryString("uid")

' run the transformation
xslProc.transform

On your stylesheet you'll to have a global 'uid' parameter

Hope this helps




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